Answer
$$
\begin{aligned}
y^{\prime}=\frac{d}{dx} \left[ y \right] &= \frac{d}{dx} \left[ \arctan (\cos \theta)\right] =-\frac{\sin \theta}{1+\cos ^{2} \theta}
\end{aligned}
$$
Work Step by Step
$$
y=\arctan (\cos \theta)
$$
Differentiating both sides of this equation we have
$$
\begin{aligned}
\frac{d}{dx} \left[ y \right] &= \frac{d}{dx} \left[ \arctan (\cos \theta)\right] \\
y^{\prime}&=\frac{1}{1+(\cos \theta)^{2}}(-\sin \theta)\\
&=-\frac{\sin \theta}{1+\cos ^{2} \theta}
\end{aligned}
$$
