Answer
$$y' =\cot x -\sin x\cos x$$
Work Step by Step
Given $$y=\ln\left(\sin x\right)-\frac{1}{2}\sin^2x$$
Then
\begin{align*}
y'&=\frac{d}{dx}\left(\ln \left(\sin \left(x\right)\right)\right)-\frac{d}{dx}\left(\frac{1}{2}\sin ^2\left(x\right)\right)\\
&=\frac{\cos x}{\sin x} -\sin x\cos x\\
&=\cot x -\sin x\cos x
\end{align*}
