I'll set $\theta=\phi$ for typographical convenience. Now, we have that $g_r=f_r.$ Also, setting $x=r\cos\phi$ and $y=r\sin\phi,$ we have that $$g_r=f_r=f_xx_r+f_yy_r.$$
Also note that when $r=\sqrt 2,\,\phi=π/4,$ we have that $(x,y)=(1,1).$
Can you now complete this?
Is my thought process correct? I feel like I may have misunderstood how the chain rule works with partial derivatives and would appreciate a sanity check for this.
The standard approach is chain rule, product rule, chain rule$\times 2$. Putting it all together you should obtain:
$$\small\begin{align}
\dfrac{\partial^2 z}{\partial x~\partial y}&=\dfrac{\partial~~}{\partial y}\left[\dfrac{\partial z}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial v}{\partial x}\right]
\\&=\dfrac{\partial z}{\partial u}\dfrac{\partial^2 u}{\partial x\,\partial y}+\dfrac{\partial~~}{\partial y}\left[\dfrac{\partial z}{\partial u}\right]\cdot\dfrac{\partial u}{\partial x}+\dfrac{\partial ~~}{\partial y}\left[\dfrac{\partial z}{\partial v}\right]\cdot\dfrac{\partial v}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial^2 v }{\partial x\,\partial y}
\\[1ex]&=\dfrac{\partial z}{\partial u}\dfrac{\partial^2 u}{\partial x\,\partial y}+\left[\dfrac{\partial^2 z }{\partial u~^2}\dfrac{\partial u}{\partial y}+\dfrac{\partial^2 z}{\partial u\,\partial v}\dfrac{\partial v}{\partial y}\right]\dfrac{\partial u}{\partial x}+\left[\dfrac{\partial^2 z}{\partial u\,\partial v }\dfrac{\partial u}{\partial y}+\dfrac{\partial^2 z}{\partial v~^2}\dfrac{\partial v}{\partial y}\right]\dfrac{\partial v}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial^2 v }{\partial x\,\partial y}
\\[1ex]&=\dfrac{\partial z}{\partial u}\dfrac{\partial^2 u}{\partial x\,\partial y}+\dfrac{\partial^2 z }{\partial u~^2}\dfrac{\partial u}{\partial y}\dfrac{\partial u}{\partial x}+\dfrac{\partial^2 z}{\partial u\,\partial v}\left[\dfrac{\partial v}{\partial y}\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial x}\dfrac{\partial u}{\partial y}\right]+\dfrac{\partial^2 z}{\partial v~^2}\dfrac{\partial v}{\partial y}\dfrac{\partial v}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial^2 v }{\partial x\,\partial y}
\\[1ex]&=\dfrac{\partial z}{\partial u}\dfrac{\partial^2 xy^2}{\partial x\,\partial y}+\dfrac{\partial^2 z}{\partial u~^2}\dfrac{\partial xy^2}{\partial y}\dfrac{\partial xy^2}{\partial x}+\dfrac{\partial^2 z}{\partial u\,\partial v}\left[\dfrac{\partial x^2}{\partial y}\dfrac{\partial xy^2}{\partial x}+\dfrac{\partial x^2}{\partial x}\dfrac{\partial xy^2}{\partial y}\right]+\dfrac{\partial^2 z}{\partial v~^2}\dfrac{\partial x^2}{\partial y}\dfrac{\partial x^2}{\partial x}+\dfrac{\partial z}{\partial v}\dfrac{\partial^2 x^2 }{\partial x\,\partial y}
\\[1ex]&=\dfrac{\partial z}{\partial u}2y+\dfrac{\partial^2 z }{\partial u~^2 }4xy^3+\dfrac{\partial^2 z}{\partial u\,\partial v}\left[0+4x^2y\right]+0+0
\\[1ex]&=2y f_u(xy^2,x^2)+2xy^3f_{uu}(xy^2,x^2)+4x^2yf_{uv}(xy^2,x^2)\end{align}$$
So...$\color{green}\checkmark$
Or more quickly: $$\begin{align}u&=xy^2\\ v&=x^2\\\dfrac{\partial^2 f(u,v)}{\partial x~\partial y} ~&=~\dfrac{\partial ~~}{\partial x}(\dfrac{\partial u}{\partial y}f_u(u,v)+\dfrac{\partial v}{\partial y} f_v(u,v)\\&=\dfrac{\partial~~}{\partial x}(2xyf_u(u,v)+0)\\&=\dfrac{\partial 2xy}{\partial x}~f_u(u,v)+(2xy)~\dfrac{\partial f_u(u,v)}{\partial x}\\&=(2y)f_u(u,v)+(2xy)(\dfrac{\partial u}{\partial x}f_{uu}(u,v)+\dfrac{\partial v}{\partial x}f_{uv}(u,v))\\&=2yf_u(xy^2,x^2)+2xy^3f_{uu}(xy^2,x^2)+4x^2yf_{uv}(xy^2,x^2)\end{align}$$
How do I compute the partial derivatives of f with respect to u and v? I have not been given any further information;
Therefore, while you have been promised that these derivatives exist, you cannot compute them.
So, that is all you can do and your job is done.
Best Answer
The equation $$F(cx - az, cy-bz) = 0$$ defines implicitly a function $(x,y)\mapsto z(x,y)$ under reasonable hypothesis (see http://en.wikipedia.org/wiki/Implicit_function_theorem). Applying the chain rule to $$\pmatrix{x\cr y}\longmapsto \pmatrix{cx-az(x,y)\cr cy-bz(x,y)}\longmapsto F(cx - az(x,y), cy-bz(x,y))$$ and using that $F(cx - az(x,y), cy-bz(x,y))=0$ forall $(x,y)$ you will obtain a system of two equations with two unknowns $\partial z/\partial x$ and $\partial z/\partial y$.