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