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