Answer
$$f(x)=\ln x+C$$
Work Step by Step
Given $$ \int f(x) e^{x} d x=f(x) e^{x}-\int x^{-1} e^{x} d x$$
Let
\begin{align*}
u&= f(x) \ \ \ \ \ \ \ \ \ \ dv= e^xdx\\
du&=f'(x) dx\ \ \ \ \ \ v=e^x
\end{align*}
Then
\begin{align*}
\int f(x) e^{x} d x&= f(x) e^x -\int f'(x)e^xdx\\
f(x) e^{x}-\int x^{-1} e^{x} d x&= f(x) e^x -\int f'(x)e^xdx
\end{align*}
By comparison
\begin{align*}
\int x^{-1} e^{x} d x&= \int f'(x)e^xdx\\
\int x^{-1} d x&= \int f'(x) dx\\
\ln x+C&=f(x)
\end{align*}
