Answer
$$\frac{{dy}}{{dt}} = \frac{9}{{{t^2}{e^{3/t}}}}$$
Work Step by Step
$$\eqalign{
& y = 3{e^{ - 3/t}} \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {3{e^{ - 3/t}}} \right] \cr
& \frac{{dy}}{{dt}} = 3{e^{ - 3/t}}\frac{d}{{dt}}\left[ { - \frac{3}{t}} \right] \cr
& {\text{Multiply and simplify}} \cr
& \frac{{dy}}{{dt}} = - 3{e^{ - 3/t}}\left( { - \frac{3}{{{t^2}}}} \right) \cr
& \frac{{dy}}{{dt}} = \frac{{9{e^{ - 3/t}}}}{{{t^2}}} \cr
& \frac{{dy}}{{dt}} = \frac{9}{{{t^2}{e^{3/t}}}} \cr} $$
