Answer
$$-e^{2}$$
Work Step by Step
Given $$f(x) =xe^{-x} $$
Since
\begin{align*}
f'(x) &= \frac{d}{dx}\left(x\right)e^{-x}+\frac{d}{dx}\left(e^{-x}\right)x\\
&= e^{-x}-e^{-x}x\\
&=e^{-x}(1-x)
\end{align*}
Since $f^{\prime}(x)<0$ for $x>1$, then the function is strictly decreasing and one-to-one. Hence, the inverse exists for $f(x)$ on $[1, \infty)$
Consider that $g$ is the inverse of $f$. Then
$$f(2)=2e^{-2}\ \ \ \ \to\ \ g( 2e^{-2}) =2 $$
and
\begin{align*}
g^{\prime}\left(2 e^{-2}\right)&=\frac{1}{f^{\prime}\left(g\left(2 e^{-2}\right)\right)}\\
&=\frac{1}{f^{\prime}(2)}\\
&=\frac{1}{e^{-2}(1-2)}=-e^{2}
\end{align*}
