Answer
$$ f'(x) =\frac{(3x-2)e^x}{(3x+1)^2}$$
Work Step by Step
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Recall that $(e^x)'=e^x$
Since we have
$$ f(x)= \frac{e^x}{3x+1}$$
then the derivative $ f'(x)$, using the chain and quotient rules, is given by
$$ f'(x)=\frac{(3x+1)(e^x)'-e^x(3x+1)'}{(3x+1)^2} =\frac{(3x+1)e^x-3e^x}{(3x+1)^2}\\
=\frac{(3x-2)e^x}{(3x+1)^2}$$
