Explanation

(d)
$\because\ \ f(x) = \log_{x}{(\log_{e}{x) = \frac{\log_{e}{\log_{e}x}}{\log_{e}x}}}$
$\therefore\ \$ $f(x) = \frac{\left\lbrack \log_{e}{x\frac{1}{\log_{e}x}.\frac{1}{x} - \log{\log_{e}{x.\frac{1}{x}}}} \right\rbrack}{{(\log_{e}{x)}}^{2}}$
$= \ \frac{1 - \log_{e}{\log_{e}x}}{x({\log_{e}{x)}}^{2}}$
$\Rightarrow \$ $f^{'}(e) = \frac{1 - \log_{e}{\log_{e}e}}{e{(\log_{e}{e)}}^{2}} = \frac{1 - \log_{e}1}{e} = \frac{1}{e}$