Answer
$A^{-1}=\left[\begin{array}{ll}
e^{-2x}/2 & -e^{-3x}/2\\
-e^{-x}/2 & e^{-2x}/2
\end{array}\right]$
Work Step by Step
If $A=\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]$, then $A^{-1}=\displaystyle \frac{1}{ad-bc}\left[\begin{array}{ll}
d & -b\\
-c & a
\end{array}\right]$.
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$ad-bc=e^{2x}e^{2x}-(-e^{x})(e^{3x})=e^{4x}+e^{4x}=2e^{4x}$
$ A^{-1}=\displaystyle \frac{1}{2e^{4x}}\left[\begin{array}{ll}
e^{2x} & e^{x}\\
-e^{3x} & e^{2x}
\end{array}\right]=\displaystyle \qquad$ ... $ \displaystyle \frac{e^{M}}{e^{N}}=e^{M-N}$
$=\left[\begin{array}{ll}
e^{2x-4x}/2 & e^{x-4x}/2\\
-e^{3x-4x}/2 & e^{2x-4x}/2
\end{array}\right]$
$A^{-1}=\left[\begin{array}{ll}
e^{-2x}/2 & -e^{-3x}/2\\
-e^{-x}/2 & e^{-2x}/2
\end{array}\right]$
Check:
$AA^{-1}=\left[\begin{array}{ll}
e^{2x} & -e^{x}\\
e^{3x} & e^{2x}
\end{array}\right]\left[\begin{array}{ll}
e^{-2x}/2 & -e^{-3x}/2\\
-e^{-x}/2 & e^{-2x}/2
\end{array}\right]$
$=\left[\begin{array}{ll}
e^{0}/2+e^{0}/2 & -e^{-x}/2+e^{-x}/2\\
-e^{-x}/2+e^{-x}/2 & e^{0}/2+e^{0}/2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]$
