Answer
$\dfrac{\partial f}{\partial \theta}=-3r \cos^2 \theta \sin \theta$
Work Step by Step
We have $\dfrac{\partial f}{\partial x}=y; \dfrac{\partial f}{\partial y}=x; \dfrac{\partial f}{\partial z}=-2z; \dfrac{\partial x}{\partial \theta}=-r \sin \theta; \dfrac{\partial y}{\partial \theta}=-2 \cos \theta \sin \theta $ and $\dfrac{\partial z}{\partial \theta}=0$
We apply the chain rule to obtain:
$\dfrac{\partial f}{\partial \theta}=\dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial \theta}+\dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial \theta}+\dfrac{\partial f}{\partial z} \dfrac{\partial y}{\partial \theta} \\=-yr \sin \theta -2x \cos \theta \sin \theta +0$
We are given that $x=r \cos \theta; y=\cos^2 \theta, z=r$
Thus, $\dfrac{\partial f}{\partial \theta}=-r \cos^2 \theta \sin \theta -2r \cos^2 \theta \sin \theta $
Therefore, we get $\dfrac{\partial f}{\partial \theta}=-3r \cos^2 \theta \sin \theta$
