My work is as follows. Criticism welcomed.
$$y = \cot^2(\sin\theta) = (\cot(\sin\theta))^2$$
Power Rule combined with the Chain Rule:
$$\begin{align} y' & = 2(\cot(\sin \theta)) \cdot \frac d{dx}(\cot(\sin\theta)) \cdot \frac d{dx}(\sin \theta) \\ \\
&= 2(\cot(\sin \theta)) \cdot (- \csc^2(\sin \theta)) \cdot \cos \theta
\end{align}$$
Best Answer
As I said in my comment above: Your work is correct.
As far as making it "elegant", I would simply pull the negative (the coefficient of $\csc^2(\sin\theta))$ to the front: $$-2\cot(\sin\theta)\csc^2(\sin\theta)(\cos \theta),$$
Other than that, you might want to bring the factor of $\cos \theta$ to the front as well: $$-2(\cos \theta) \cot(\sin \theta)\csc^2(\sin\theta).$$
Otherwise, elegance is in the eye of the beholder!