Answer
$$ f'(x)=\frac{(1+x )e^{-x} }{x^2}.$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Since $ f(x)= \frac{e^{-x}}{x}$, then the derivative, by using the quotient rule, is given by
$$ f'(x)=\frac{e^{-x}-x e^{-x} (-x)'}{x^2}=\frac{e^{-x}+x e^{-x} }{x^2}=\frac{(1+x )e^{-x} }{x^2}.$$
