Answer
$$ f'(t)= -\frac{ 3e^{-3t}}{(1-e^{-3t})^2}.$$
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
$$ f(t)= \frac{1}{1-e^{-3t}}$$
then the derivative $ f'(t)$, using the chain and quotient rules, is given by
$$ f'(t)= \frac{(1-e^{-3t})(1)'-(1-e^{-3t})'}{(1-e^{-3t})^2}=\frac{ -(0-e^{-3t}(-3))}{(1-e^{-3t})^2}\\
=-\frac{ 3e^{-3t}}{(1-e^{-3t})^2}.$$
