Answer
$$\frac{{dy}}{{d\theta }} = \frac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{2\sin \theta \cos \theta }} - \frac{2}{{\theta \left( {1 + 2\ln \theta } \right)}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\frac{{\sqrt {\sin \theta \cos \theta } }}{{1 + 2\ln \theta }}} \right) \cr
& {\text{use the quotient property for logarithms}} \cr
& y = \ln \left( {\sqrt {\sin \theta \cos \theta } } \right) - \ln \left( {1 + 2\ln \theta } \right) \cr
& {\text{rewrite the radical}} \cr
& y = \ln {\left( {\sin \theta \cos \theta } \right)^{1/2}} - \ln \left( {1 + 2\ln \theta } \right) \cr
& {\text{use the property for logarithms}} \cr
& y = \frac{1}{2}\ln \left( {\sin \theta \cos \theta } \right) - \ln \left( {1 + 2\ln \theta } \right) \cr
& {\text{find the derivative of }}y{\text{ with respect to }}\theta \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{2}\frac{d}{{d\theta }}\left[ {\ln \left( {\sin \theta \cos \theta } \right)} \right] - \frac{d}{{d\theta }}\left[ {\ln \left( {1 + 2\ln \theta } \right)} \right] \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\sin \theta \cos \theta }}\frac{d}{{d\theta }}\left[ {\sin \theta \cos \theta } \right] - \frac{d}{{d\theta }}\left[ {\ln \left( {1 + 2\ln \theta } \right)} \right] \cr
& {\text{use the product rule for }}\frac{d}{{d\theta }}\left[ {\ln \left( {\sin \theta \cos \theta } \right)} \right] \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\sin \theta \cos \theta }}\left( {\sin \theta \frac{d}{{d\theta }}\left[ {\cos \theta } \right] + \cos \theta \frac{d}{{d\theta }}\left[ {\sin \theta } \right]} \right) - \frac{d}{{d\theta }}\left[ {\ln \left( {1 + 2\ln \theta } \right)} \right] \cr
& {\text{solve the derivatives}} \cr
& \frac{{dy}}{{d\theta }} = \frac{1}{{2\sin \theta \cos \theta }}\left( { - {{\sin }^2}\theta + {{\cos }^2}\theta } \right) - \frac{1}{{1 + 2\ln \theta }}\left( {\frac{2}{\theta }} \right) \cr
& {\text{simplifying, we get:}} \cr
& \frac{{dy}}{{d\theta }} = \frac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{2\sin \theta \cos \theta }} - \frac{2}{{\theta \left( {1 + 2\ln \theta } \right)}} \cr} $$
