Answer
$h'(e)=\frac{1}{e+1}$
Work Step by Step
$h'(x)=(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$
$h'(e)=\frac{1}{f'(f^{-1}(e))}$
So if $f^{-1}(e)=a$ then $e=f(a)$.
$e=f(a) \to e=e^{a}+\ln a$
The above equation has a trivial solution at $a=1$ because $e=e^{1}+\ln 1 \to e=e+0 \to e=e$
So $f^{-1}(e)=a \to f^{-1}(e)=1$
$h'(e)=\frac{1}{f'(f^{-1}(e))}\to h'(e)=\frac{1}{f'(1)}$
So the derivative of $f$ is:
$f'(x)=(e^{x})'+(\ln x)' \to f'(x)=e^{x}+\frac{1}{x} \to f'(1)=e+\frac{1}{1}=e+1$
$h'(e)=\frac{1}{f'(1)} \to h'(e)=\frac{1}{e+1}$
