Inverse of an invertible upper triangular matrix of order 3
There is a nice trick for calculating the inverse of any invertible upper triangular matrix, one which avoids the computation of complicated determinants. Since it works for any such upper (or lower) triangular matrix $T$ of any size $n$, I'll explain it in that context.
The first thing one needs to remember is the determinant of a triangular matrix is the product of its diagonal entries \[\det T=\prod_{i=1}^nt_{ii}.\tag7\] It follows that the characteristic polynomial $p_T(λ)$ of $T$ is $$p_T(λ) = \det(T - λ I) = \prod_{i=1}^n (t_{ii} - λ), \tag{8}$$ and from (8) that the eigenvalues of $T$ are precisely its diagonal entries, i.e. the $t_{ii}$, $1 \le i \le n$
Also follows from (7) the related fact that $T$ is nonsingular, that is, $\det(T) \ne 0$, precisely when its diagonal entries are all nonzero.
For non-singular $T$ we may compute $T^{-1}$ as follows: write $$T = λ + T_u, \tag{9}$$ where $λ$ is the diagonal matrix formed from the diagonal of $T$; viz. $$λ = [\delta_{ij} t_{ij}]; \tag{10}$$ then $λ$ is nonsingular and $T_u = T - λ$ is the strictly upper triangular matrix obtained by setting the diagonal of $T$ to zero, i.e. setting $t_{ii} = 0$ for $1 \le i \le n$. We may write $$T = λ (I + λ^{-1} T_u), \tag{11}$$ whence $$T^{-1} = (I + λ^{-1} T_u)^{-1} λ^{-1}. \tag{12}$$ The matrix $λ^{-1} T_u$ occurring in (12) is itself in fact strictly upper triangular as well as is $T_u$; indeed, for any diagonal $D$, $DT_u$ is strictly upper triangular, an assertion which is easily validated by direct calculation. It follows that $λ^{-1} T_u$ is in fact nilpotent; that is, $(λ^{-1} T_u)^n = 0$. We may now use the well-known algebraic identity $$(1 + x)\sum_{j=0}^m (-x)^j = 1 - (-x)^{m + 1}, $$ easily seen to hold in any unital ring, applied to the matrix $x =λ^{-1} T_u$, yielding, with $m = n - 1$, $$(I + λ^{-1}T_u)\sum_{j=0}^m (-λ^{-1}T_u)^j = I - (-λ^{-1}T_u)^{m + 1} = I - (-λ^{-1}T_u)^n = I. \tag{13}$$ (13) shows that the inverse of $I + λ^{-1}T_u$ is given by $$(I + λ^{-1} T_u)^{-1} = \sum_{j=0}^m (-λ^{-1}T_u)^j. \tag{14}$$ It follows from (14) that $(I + λ T_u)^{-1}$ is upper triangular, since each of the matrices $(-λ^{-1}T_u)^j$, $j≥1$, is strictly upper triangular, and $(-λ^{-1}T_u)^0 = I$. It further follows then that $T^{-1} = (I + λ T_u)^{-1}λ^{-1}$ is also upper triangular, being the product of the upper triangular matrix $(I + λ T_u)^{-1}$ and the diagonal matrix $λ^{-1}$. We have thus shown that the inverse of any invertible upper triangular matrix, of any size $n$, is itself an upper triangular matrix.
The inverse of any invertible matrix is invertible, the inverse of the inverse being the original matrix.
We can apply these considerations to the calculation of $A^{-1}$, where $$A = \begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{bmatrix}$$ here we have $$λ = \begin{bmatrix} a & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & f \end{bmatrix}$$ and $$T_u = \begin{bmatrix} 0 & b & c \\ 0 & 0 & e \\ 0 & 0 & 0 \end{bmatrix}$$ then \begin{align*} λ^{-1} T_u &= \begin{bmatrix} a^{-1} & 0 & 0 \\ 0 & d^{-1} & 0 \\ 0 & 0 & f^{-1} \end{bmatrix} \begin{bmatrix} 0 & b & c \\ 0 & 0 & e \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix}\\ (λ^{-1} T_u)^2&= \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & bea^{-1}d^{-1} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ (λ^{-1} T_u)^3&=0\\ \sum_{j=0}^2 (-λ^{-1} T_u)^j&= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & bea^{-1}d^{-1} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ &= \begin{bmatrix} 1 & -ba^{-1} & (be - cd)a^{-1}d^{-1} \\ 0 & 1 &- ed^{-1} \\ 0 & 0 & 1 \end{bmatrix}; \end{align*} finally, \begin{align*}T^{-1}&= (I + λ^{-1} T_u)^{-1} λ^{-1}\\&= \sum_{j=0}^2 (-λ^{-1} T_u)^j λ^{-1}\\ &= \begin{bmatrix} 1 & -ba^{-1} & (be - cd)a^{-1}d^{-1} \\ 0 & 1 &- ed^{-1} \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a^{-1} & 0 & 0 \\ 0 & d^{-1} & 0 \\ 0 & 0 & f^{-1} \end{bmatrix}\\ &= \begin{bmatrix} a^{-1} & -ba^{-1}d^{-1} & (be - cd)a^{-1}d^{-1}f^{-1} \\ 0 & d^{-1} &- ed^{-1}f^{-1} \\ 0 & 0 & f^{-1} \end{bmatrix} \end{align*}
Also see
Use of formal series to compute inverse in abstract rings
Strictly upper triangular matrices are nilpotent
There is a nice trick for calculating the inverse of any invertible upper triangular matrix, one which avoids the computation of complicated determinants. Since it works for any such upper (or lower) triangular matrix $T$ of any size $n$, I'll explain it in that context.
The first thing one needs to remember is the determinant of a triangular matrix is the product of its diagonal entries \[\det T=\prod_{i=1}^nt_{ii}.\tag7\] It follows that the characteristic polynomial $p_T(λ)$ of $T$ is $$p_T(λ) = \det(T - λ I) = \prod_{i=1}^n (t_{ii} - λ), \tag{8}$$ and from (8) that the eigenvalues of $T$ are precisely its diagonal entries, i.e. the $t_{ii}$, $1 \le i \le n$
Also follows from (7) the related fact that $T$ is nonsingular, that is, $\det(T) \ne 0$, precisely when its diagonal entries are all nonzero.
For non-singular $T$ we may compute $T^{-1}$ as follows: write $$T = λ + T_u, \tag{9}$$ where $λ$ is the diagonal matrix formed from the diagonal of $T$; viz. $$λ = [\delta_{ij} t_{ij}]; \tag{10}$$ then $λ$ is nonsingular and $T_u = T - λ$ is the strictly upper triangular matrix obtained by setting the diagonal of $T$ to zero, i.e. setting $t_{ii} = 0$ for $1 \le i \le n$. We may write $$T = λ (I + λ^{-1} T_u), \tag{11}$$ whence $$T^{-1} = (I + λ^{-1} T_u)^{-1} λ^{-1}. \tag{12}$$ The matrix $λ^{-1} T_u$ occurring in (12) is itself in fact strictly upper triangular as well as is $T_u$; indeed, for any diagonal $D$, $DT_u$ is strictly upper triangular, an assertion which is easily validated by direct calculation. It follows that $λ^{-1} T_u$ is in fact nilpotent; that is, $(λ^{-1} T_u)^n = 0$. We may now use the well-known algebraic identity $$(1 + x)\sum_{j=0}^m (-x)^j = 1 - (-x)^{m + 1}, $$ easily seen to hold in any unital ring, applied to the matrix $x =λ^{-1} T_u$, yielding, with $m = n - 1$, $$(I + λ^{-1}T_u)\sum_{j=0}^m (-λ^{-1}T_u)^j = I - (-λ^{-1}T_u)^{m + 1} = I - (-λ^{-1}T_u)^n = I. \tag{13}$$ (13) shows that the inverse of $I + λ^{-1}T_u$ is given by $$(I + λ^{-1} T_u)^{-1} = \sum_{j=0}^m (-λ^{-1}T_u)^j. \tag{14}$$ It follows from (14) that $(I + λ T_u)^{-1}$ is upper triangular, since each of the matrices $(-λ^{-1}T_u)^j$, $j≥1$, is strictly upper triangular, and $(-λ^{-1}T_u)^0 = I$. It further follows then that $T^{-1} = (I + λ T_u)^{-1}λ^{-1}$ is also upper triangular, being the product of the upper triangular matrix $(I + λ T_u)^{-1}$ and the diagonal matrix $λ^{-1}$. We have thus shown that the inverse of any invertible upper triangular matrix, of any size $n$, is itself an upper triangular matrix.
The inverse of any invertible matrix is invertible, the inverse of the inverse being the original matrix.
We can apply these considerations to the calculation of $A^{-1}$, where $$A = \begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f \end{bmatrix}$$ here we have $$λ = \begin{bmatrix} a & 0 & 0 \\ 0 & d & 0 \\ 0 & 0 & f \end{bmatrix}$$ and $$T_u = \begin{bmatrix} 0 & b & c \\ 0 & 0 & e \\ 0 & 0 & 0 \end{bmatrix}$$ then \begin{align*} λ^{-1} T_u &= \begin{bmatrix} a^{-1} & 0 & 0 \\ 0 & d^{-1} & 0 \\ 0 & 0 & f^{-1} \end{bmatrix} \begin{bmatrix} 0 & b & c \\ 0 & 0 & e \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix}\\ (λ^{-1} T_u)^2&= \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & bea^{-1}d^{-1} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ (λ^{-1} T_u)^3&=0\\ \sum_{j=0}^2 (-λ^{-1} T_u)^j&= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & ba^{-1} & ca^{-1} \\ 0 & 0 & ed^{-1} \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & bea^{-1}d^{-1} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ &= \begin{bmatrix} 1 & -ba^{-1} & (be - cd)a^{-1}d^{-1} \\ 0 & 1 &- ed^{-1} \\ 0 & 0 & 1 \end{bmatrix}; \end{align*} finally, \begin{align*}T^{-1}&= (I + λ^{-1} T_u)^{-1} λ^{-1}\\&= \sum_{j=0}^2 (-λ^{-1} T_u)^j λ^{-1}\\ &= \begin{bmatrix} 1 & -ba^{-1} & (be - cd)a^{-1}d^{-1} \\ 0 & 1 &- ed^{-1} \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a^{-1} & 0 & 0 \\ 0 & d^{-1} & 0 \\ 0 & 0 & f^{-1} \end{bmatrix}\\ &= \begin{bmatrix} a^{-1} & -ba^{-1}d^{-1} & (be - cd)a^{-1}d^{-1}f^{-1} \\ 0 & d^{-1} &- ed^{-1}f^{-1} \\ 0 & 0 & f^{-1} \end{bmatrix} \end{align*}
Also see
Use of formal series to compute inverse in abstract rings
Strictly upper triangular matrices are nilpotent