If $f(x)={\log }_x(\log x)$, then $f^{\prime }(x)$ at $x=e$ is:

$f(x)={\log }_x(\log x)=\frac{\ln (\ln x)}{\ln x}$ (using base change formula).Differentiating using the quotient rule ${\left(\frac{u}{v}\right)}^{\prime }=\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$:$f^{\prime }(x)=\frac{(\ln x)\cdot \frac{d}{dx}[\ln (\ln x)]-\ln (\ln x)\cdot \frac{d}{dx}[\ln x]}{(\ln x{)}^2}$$f^{\prime }(x)=\frac{(\ln x)\cdot (\frac{1}{\ln x}\cdot \frac{1}{x})-\ln (\ln x)\cdot \frac{1}{x}}{(\ln x{)}^2}$$f^{\prime }(x)=\frac{\frac{1}{x}-\frac{\ln (\ln x)}{x}}{(\ln x{)}^2}$ At $x=e$:$f^{\prime }(e)=\frac{\frac{1}{e}-\frac{\ln (\ln e)}{e}}{(\ln e{)}^2}$ Since $\ln e=1$ and $\ln 1=0$:$f^{\prime }(e)=\frac{\frac{1}{e}-\frac{0}{e}}{1^2}=\frac{1}{e}$