Answer
$A^{-1}=\left[\begin{array}{ll}
e^{-x}/2 & -e^{-3x}/2\\
e^{-3x}/2 & e^{-5x}/2
\end{array}\right]$
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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$ad-bc=e^{x}e^{5x}-e^{3x}(-e^{3x})=e^{6x}+e^{6x}=2e^{6x}$
$ A^{-1}=\displaystyle \frac{1}{2e^{6x}}\left[\begin{array}{ll}
e^{5x} & -e^{3x}\\
e^{3x} & e^{x}
\end{array}\right]=\displaystyle \qquad$ ... $ \displaystyle \frac{e^{M}}{e^{N}}=e^{M-N}$
$=\left[\begin{array}{ll}
e^{5x-6x}/2 & -e^{3x-6x}/2\\
e^{3x-6x}/2 & e^{x-6x}/2
\end{array}\right]$
$A^{-1}=\left[\begin{array}{ll}
e^{-x}/2 & -e^{-3x}/2\\
e^{-3x}/2 & e^{-5x}/2
\end{array}\right]$
Check:
$AA^{-1}=\left[\begin{array}{ll}
e^{x} & e^{3x}\\
-e^{3x} & e^{5x}
\end{array}\right]\left[\begin{array}{ll}
e^{-x}/2 & -e^{-3x}/2\\
e^{-3x}/2 & e^{-5x}/2
\end{array}\right]$
$=\left[\begin{array}{ll}
e^{0}/2+e^{0}/2 & -e^{-2x}/2+e^{-2x}/2\\
-e^{-2x}/2+e^{-2x}/2 & e^{0}/2+e^{0}/2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]$