Evaluating $\frac{dg}{dθ}$ at $(r,θ)=(2\sqrt{2},\frac{π}{4})$ where $g(x,y)=\frac1{x+y^2}$ using chain rule? 
Okay so my first step is to find the partial derivatives:
$$\frac{\partial \:}{\partial \:x}\left(\frac{1}{x+y^2}\right)=-\frac{1}{\left(x+y^2\right)^2}$$
$$\frac{\partial \:}{\partial \:y}\left(\frac{1}{x+y^2}\right)=-\frac{2y}{\left(x+y^2\right)^2}$$
Then I multiply each partial derivative with its respective equal and add it together:
$$(fx)(32r\cos(\theta)) + (fy)(3r(\sin(\theta))$$
Then I sub in X and Y in the partial derivatives. And then plug in $r$ and $\theta$:
$$-\left(\frac{1}{\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:}{4}\right)+\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi }{4}\right)\right)^2\right)^2}\right)\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:}{4}\right)\right)+-\left(\frac{2\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)}{\left(32\left(2\sqrt{2}\right)cos\left(\frac{\pi \:\:}{4}\right)+\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)^2\right)^2}\right)\left(3\left(2\sqrt{2}\right)sin\left(\frac{\pi \:}{4}\right)\right)$$
And I show my answer in the screenshot above. But for some reason the software is telling me it's wrong. I've also tried not multiplying the partial derivatives by the $32rcos(\theta)$ thing but it's still giving me the wrong answer.
What am I doing wrong? Thank you.
 A: It is easier if you substitute $x$ and $y$ directly $$g(r,\theta)=\frac{1}{x+y^2}=\frac{1}{32r\cos(\theta)+9r^2\sin^2(\theta)}$$
So $$\frac{\partial g}{\partial \theta}=-\frac{-32r\sin(\theta)+18r^2\sin(\theta)\cos(\theta)}{(32r\cos(\theta)+9r^2\sin^2(\theta))^2}.$$

Or $$\frac{\partial g}{\partial \theta}=\frac{\partial g }{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial g }{\partial y}\frac{\partial y}{\partial \theta}$$
$$=-\frac{1}{(x+y^2)^2}(-32r\sin(\theta)+2y(3r\cos(\theta)))$$
$$=\frac{32r\sin(\theta)-18r^2\sin(\theta)\cos(\theta)}{(32r\cos(\theta)+9r^2\sin^2(\theta))^2}.$$
Then evaluating at $(r,\theta)=(2\sqrt{2},\frac{\pi}{4})$ you should get $-\frac{1}{1250}$.
A: We have that
$$\frac{\partial g}{\partial \theta}=\frac{\partial g}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial g}{\partial y}\frac{\partial y}{\partial \theta}=-\frac{1}{\left(x+y^2\right)^2}(-32 r \sin \theta)-\frac{2y}{\left(x+y^2\right)^2}(3r\cos \theta)=$$
with $x=64$ and $y=6$ then
$$=-\frac{1}{100^2}(-64)-\frac{12}{100^2}(6)=-\frac{8}{100^2}$$
