#### Answer

$$\frac{-1}{6}$$

#### Work Step by Step

Given $$g(x, y)=\frac{1}{x+y^{2}},\ \ \ \ \ x=r\cos\theta,\ \ \ y=r\sin \theta ,\ \ (r,\theta )= \left(2 \sqrt{2}, \frac{\pi}{4}\right)$$
We see that at $(r,\theta )= \left(2 \sqrt{2}, \frac{\pi}{4}\right)$, we have $(x,y)= (2,2)$ and \begin{align*} \frac{\partial g }{\partial \:x}&= \frac{-1}{\left(x+y^2\right)^2},\ \ \ \ \ \frac{\partial g }{\partial \:y}=\frac{-2y}{\left(x+y^2\right)^2}\\ \frac{\partial x }{\partial \:\theta}&=-r\sin \theta,\ \ \ \ \ \ \ \ \ \ \frac{\partial y }{\partial \:\theta}=r\cos\theta \end{align*} Then \begin{align*} \frac{\partial g }{\partial \:\theta }&= \frac{\partial g }{\partial \:x} \frac{\partial x }{\partial \:\theta}+ \frac{\partial g }{\partial \:y} \frac{\partial y }{\partial \:\theta}\\ &=\frac{r\sin \theta}{(x+y^2)^2} +\frac{2r\cos \theta}{(x+y^2)^2}\\ &=\frac{y}{(x+y^2)^2} -\frac{2x}{(x+y^2)^2} \end{align*} Hence $$ \frac{\partial g }{\partial \:\theta }\bigg|_{(x,y)=(2,2)}=\frac{-1}{6}$$