Answer
$\begin{array}( \\\frac{1}{2} \\ \\ \end{array}
\begin{bmatrix} 1 &e^{-x} &0\\e^{-x} &-e^{-2x}&0\\0 &0&1 \end{bmatrix} $
for all real $x$.
Work Step by Step
Step 1. Write the matrix together with an identify as:
$\begin{array}( \\A|I= \\ \\ \end{array}
\begin{bmatrix} 1 & e^x &0 &| &1 &0 &0\\e^x &-e^{2x}&0 &| &0 &1&0\\0 &0&2 &| &0 &0&1 \end{bmatrix} \begin{array}( \\R_1e^x-R_2\to R_2 \\ \\ \end{array}$
Step 2. Use row operations to transform the left half into a reduced row-echelon:
$\begin{bmatrix} 1 & e^x &0 &| &1 &0 &0\\0 &2e^{2x}&0 &| &e^x &-1&0\\0 &0&2 &| &0 &0&1 \end{bmatrix} \begin{array}( R_1-R_2/(2e^x)\to R_1 \\ R_2/2e^{2x}\\ R_3/2\\ \end{array}$
$\begin{bmatrix} 1 & 0 &0 &| &1/2 &e^{-x}/2 &0\\0 &1&0 &| &e^{-x}/2 &-e^{-2x}/2&0\\0 &0&1 &| &0 &0&1/2 \end{bmatrix} $
Step 3. The inverse of the original matrix is:
$\begin{array}( \\A^{-1}=\frac{1}{2} \\ \\ \end{array}
\begin{bmatrix} 1 &e^{-x} &0\\e^{-x} &-e^{-2x}&0\\0 &0&1 \end{bmatrix} $
Apparently, the inverse is valid for all real x values.
