$f\left(x\right)=x^x\Rightarrow \log f\left(x\right)=x\log x$
Differentiating with respect to $x$, we get,

 $f^{\prime }\left(x\right)=x^x\left(1+\log x\right)$
Now, by property $g^{\prime }\left(f\left(x\right)\right)=\frac{1}{f^{\prime }\left(x\right)}$ {$\because g\left(x\right)$ is the inverse of $f\left(x\right)$}
Putting $x=2$
$$\begin{gathered} g^{\text{'}}\left(f\left(2\right)\right)=\frac{1}{f^{\text{'}}\left(2\right)} \\ \Rightarrow g^{\text{'}}\left(4\right)=\frac{1}{2^2\left(1+\log 2\right)}=\frac{1}{4}\left({\log }_{2e}e\right) \end{gathered}$$