Answer
$$\frac{{dy}}{{dt}} = \frac{{1 - t}}{t}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {3t{e^{ - t}}} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\ln \left( {3t{e^{ - t}}} \right)} \right] \cr
& {\text{use the rule }}\frac{d}{{dt}}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dt}};{\text{ for this exercise consider }}u = 3t{e^{ - t}}.{\text{ then}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{3t{e^{ - t}}}}\frac{d}{{dt}}\left[ {3t{e^{ - t}}} \right] \cr
& {\text{use the product rule }} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{3t{e^{ - t}}}}\left( {3t\frac{d}{{dt}}\left[ {{e^{ - t}}} \right] + {e^{ - t}}\frac{d}{{dt}}\left[ {3t} \right]} \right) \cr
& {\text{solving derivatives }} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{3t{e^{ - t}}}}\left( {3t\left( { - {e^{ - t}}} \right) + {e^{ - t}}\left( 3 \right)} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dt}} = \frac{1}{{3t{e^{ - t}}}}\left( {3{e^{ - t}} - 3t{e^{ - t}}} \right) \cr
& \frac{{dy}}{{dt}} = \frac{{3{e^{ - t}}\left( {1 - t} \right)}}{{3t{e^{ - t}}}} \cr
& \frac{{dy}}{{dt}} = \frac{{1 - t}}{t} \cr} $$
