Answer
$$ f'(x)= \frac{ e^x-4}{e^x-4x}.$$
Work Step by Step
Recall that $(\ln x)'=\dfrac{1}{x}$
Recall that $(e^x)'=e^x$
Since $ f(x)= \ln(e^x-4x)$, then the derivative, using the chain rule, is given by
$$ f'(x)= \frac{ 1}{e^x-4x}(e^x-4x)'= \frac{ e^x-4}{e^x-4x}.$$
