$f(x)=\log _{x^2}(\log x)$
$\therefore f^{\prime}(x)=\frac{d}{d x} \log _{x^2}(\log x)$
$=\frac{d}{d x}\left\{\frac{\log (\log x)}{\log \left(x^2\right)}\right\}$
$\left[\because \frac{d}{d x} \frac{u}{v}=\frac{v \cdot u^{\prime}-u v^{\prime}}{v^2}, \frac{d}{d x} \log x=\frac{1}{x}\right.$ and $\left.\log _a b=\frac{\log _e b}{\log _e a}\right]$
$\begin{aligned} & =\frac{d}{d x}\left\{\frac{\log (\log x)}{2 \log x}\right\} \\ & =\frac{1}{2}\left\{\frac{\log x \cdot \frac{1}{\log x} \cdot \frac{1}{x}-\log (\log x) \cdot \frac{1}{x}}{(\log x)^2}\right\} \\ & =\frac{1}{2}\left[\frac{1}{x} \frac{(1-\log (\log x)}{(\log x)^2}\right]\end{aligned}$
$\begin{aligned} & \text { At, } x=e \\ & f^{\prime}(e)=\frac{1}{2 e\left(\log _e e\right)^2}\left(1-\log _e\left(\log _e e\right)\right) \\ & f^{\prime}(e)=\frac{1}{2 e \cdot(1)^2}(1-0)=\frac{1}{2 e}=(2 e)^{-1}\end{aligned}$