If $f(x)=\log _x\{\ln (x)\}$, then f'(x) at x = e is

$e^{-1}$

We have,

$f(x)=\log _x\{\ln (x)\}=\frac{\ln (\ln (x))}{\ln (x)}$

∴   $f'(x)=\frac{\ln (x) . \frac{1}{\ln (x)} . \frac{1}{x}-\ln \{\ln (x)\} \frac{1}{x}}{(\ln (x))^2}=\frac{1-\ln \{\ln (x)\}}{x\{\ln (x)\}^2}$

$\Rightarrow f'(e)=\frac{1-\ln \{\ln (e))\}}{e\{\ln (e)\}^2}=\frac{1-\ln (1)}{e}=\frac{1}{e}$