If \(\mathrm{f}(x)=\log _{x^2}\left(\log _{\mathrm{e}} x\right)\), then \(\mathrm{f}^{\prime}(x)\) at \(x=\mathrm{e}\) is

\(\begin{aligned} f(x) & =\log _{x^2}\left(\log _{\mathrm{e}} x\right) \\ & =\frac{\log \left(\log _{\mathrm{e}} x\right)}{\log x^2} \\ & =\frac{\log \left(\log _{\mathrm{e}} x\right)}{2 \log x}\end{aligned}\)
\(\therefore \quad \mathrm{f}^{\prime}(x)=\frac{1}{2}\left[\frac{\log x \cdot \frac{1}{\log _{\mathrm{e}} x} \cdot \frac{\mathrm{~d}}{\mathrm{~d} x}(\log x)-\log \left(\log _{\mathrm{e}} x\right) \cdot \frac{1}{x}}{(\log x)^2}\right]\)
\(\therefore \quad \mathrm{f}^{\prime}(x)=\frac{1}{2}\left[\frac{\frac{1}{x}-\frac{\log \left(\log _{\mathrm{e}} x\right)}{x}}{(\log x)^2}\right]\)
\(\therefore \quad \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{2}\left[\frac{\frac{1}{\mathrm{e}}-0}{(1)^2}\right]=\frac{1}{2 \mathrm{e}}\)