Answer
$$\frac{{dy}}{{d\theta }} = 2\cot \theta $$
Work Step by Step
$$\eqalign{
& y = \ln \left( {{{\sin }^2}\theta } \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {\ln \left( {{{\sin }^2}\theta } \right)} \right] \cr
& {\text{use the rule }}\frac{d}{{d\theta }}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{d\theta }}\cr
&{\text{for this exercise, consider }}u = {\sin ^2}\theta :{\text{}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{{{\sin }^2}\theta }}\frac{d}{{d\theta }}\left[ {{{\sin }^2}\theta } \right] \cr
& {\text{use the chain rule}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{{{\sin }^2}\theta }}\left( {2\sin \theta } \right)\frac{d}{{d\theta }}\left[ {\sin \theta } \right] \cr
& {\text{solving the derivative, we get: }} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{{{\sin }^2}\theta }}\left( {2\sin \theta } \right)\left( {\cos \theta } \right) \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{\sin \theta }}\left( 2 \right)\left( {\cos \theta } \right) \cr
& \frac{{dy}}{{d\theta }} = 2\cot \theta \cr} $$
