$$f(x(z,w),y(z,w),z,w)=g(z,w)$$
Don't confuse $\frac{\partial f}{\partial z}$ with $\frac{\partial g}{\partial z}$
$\frac{\partial f}{\partial z}$ means the partial derivative of $f$ with respect to $z$ for $x$ independant of $z$, which is not the same for $g$.
Thus the correct expression is :
$$\frac{\partial g}{\partial z} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial z}+ \frac{\partial f}{\partial z}$$
It is clear that $$\frac{\partial g}{\partial z} \neq \frac{\partial f}{\partial z}$$
However, loosely the same symbol $f$ is used instead of two distinct symbols $f$ and $g$. This is acceptable for people familiar with this ambiguous writting. They are aware of apparent contradictions which can arise and they avoid mistakes in mentally making the distinction between the two different signification of the symbol $f$.
IN ADDITION :
Since this seems difficult to well understand, consider a concret example :
$$f(x,y,z,w)=5x+4y+3z+2w$$
$$\frac{\partial f}{\partial x}=5\quad;\quad \frac{\partial f}{\partial y}=4\quad;\quad \frac{\partial f}{\partial z}=3$$
Then consider another function $g(z,w)$ defined from the preceeding function in the particular case of
$$x(z,w)=6z+7w\quad\text{and}\quad y(z,w)=8z+9w$$
$$\frac{\partial x}{\partial z}=6\quad;\quad \frac{\partial y}{\partial z}=8$$
$$g(z,w)=5(6z+7w)+4(8z+9w)+3z+2w$$
$$g(z,w)=65z+73w$$
Obviously $5x+4y+3z+2w$ is something else than $65z+73w$ . So $f(x,y,z,w)$ and $g(z,w)$ are not a same function. They cannot be confused.
$$\frac{\partial g}{\partial z}=65$$
We see that $\frac{\partial g}{\partial z}$ is obtained directly by partial differentiation of $g(z,w)$. But instead of, if we want to compute $\frac{\partial g}{\partial z}$ indirectly from the properties of the functions $f(x,y,z,w)$ , $x(z,w)$ and $y(z,w)$ we apply the chain rule of derivations :
$$\frac{\partial g}{\partial z} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial z}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial z}+ \frac{\partial f}{\partial z}$$
$$\frac{\partial g}{\partial z} =5*6+4*8+3=65$$
As expected, the same result as before $65$ is obtained without expressing explicitely $g(z,w)$. This is purely a consequence of the chain rule. There is no need for more explanation insofar the theory of the chain rule was studied and understood.
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
If I understood your question correctly, it may help clarifying what is done right in the first step.
Note that the derivative of $z$ w.r.t. to $y$ is a function, i.e.
$$\frac{\partial z}{\partial y}(u, v)$$
makes sense and we can rename it to, say, $f(u,v) = \frac{\partial z}{\partial y}(u,v) = x^2 \frac{\partial z}{\partial u} + 2\frac{\partial z}{\partial v}$.
Now let me derive $f$ w.r.t. to $x$:
$$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y} \right) = \frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} + 2\frac{\partial z}{\partial v} \right)$$
Now what was done was, first distribute the derivative w.r.t. $x$:
$$\frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} + 2\frac{\partial z}{\partial v} \right) = \frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} \right) + \frac{\partial}{\partial x}\left(2\frac{\partial z}{\partial v} \right)$$
Now "taking $x$ out" is really the product rule $(ab)' = a'b + ab'$:
$$\frac{\partial}{\partial x}\left(x^2 \frac{\partial z}{\partial u} \right) = \frac{\partial}{\partial x}(x^2)\times \frac{\partial z}{\partial u} + x^2 \times \frac{\partial}{\partial x}\frac{\partial z}{\partial u} = 2x\frac{\partial z}{\partial u} + x^2 \frac{\partial}{\partial x}\frac{\partial z}{\partial u}$$
From there we just use the chain rule to find $ \frac{\partial}{\partial x}\frac{\partial z}{\partial u}$.
Is it clearer now?