I have a function $f(x,y) = g(u(x,y), v(x,y))$ and I need to find $\frac{\partial ^2f}{\partial x\partial y}$
I tried to solve it but I'm not too sure I'm doing it correctly. My method was the following: $$ \frac{\partial ^2f}{\partial x\partial y} = \frac{\partial }{\partial x}(\frac{\partial f}{\partial y}) $$
$$ \frac{\partial f}{\partial y} = \frac{\partial g}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial g}{\partial v}\frac{\partial v}{\partial y} $$
$$ \frac{\partial }{\partial x}(\frac{\partial g}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial g}{\partial v}\frac{\partial v}{\partial y}) = \frac{\partial }{\partial x}(\frac{\partial g}{\partial u}\frac{\partial u}{\partial y}) + \frac{\partial }{\partial x}(\frac{\partial g}{\partial v}\frac{\partial v}{\partial y}) $$
$$ \frac{\partial }{\partial x}(\frac{\partial g}{\partial u}\frac{\partial u}{\partial y}) = \frac{\partial ^2g}{\partial x\partial u}\frac{\partial u}{\partial y} + \frac{\partial g}{\partial u}\frac{\partial ^2u}{\partial x\partial y} $$
$$ \frac{\partial }{\partial x}(\frac{\partial g}{\partial v}\frac{\partial v}{\partial y}) = \frac{\partial ^2g}{\partial x\partial v}\frac{\partial v}{\partial y} + \frac{\partial g}{\partial v}\frac{\partial ^2v}{\partial x\partial y} $$ So $$ \frac{\partial ^2f}{\partial x\partial y} = \frac{\partial ^2g}{\partial x\partial u}\frac{\partial u}{\partial y} + \frac{\partial g}{\partial u}\frac{\partial ^2u}{\partial x\partial y} + \frac{\partial ^2g}{\partial x\partial v}\frac{\partial v}{\partial y} + \frac{\partial g}{\partial v}\frac{\partial ^2v}{\partial x\partial y} $$
I feel like I'm messing up at the $\frac{\partial }{\partial x}(\frac{\partial g}{\partial u}\frac{\partial u}{\partial y})$ and $\frac{\partial }{\partial x}(\frac{\partial g}{\partial v}\frac{\partial v}{\partial y})$ by using the product rule here, but I'm not sure what else I should or could do. I dont see how I'd have to use the chain rule here again if that's the intended method.
The missing detail is that you need to keep track of the points where you are computing the derivatives.
For instance, your computation of $\frac{\partial f}{\partial y}$ would be written more precisely as follows: $$\frac{\partial f}{\partial y}(x,y) = \frac{\partial g}{\partial u}(u(x,y),v(x,y))\frac{\partial u}{\partial y}(x,y) + \frac{\partial g}{\partial v}(u(x,y),v(x,y))\frac{\partial v}{\partial y}(x,y).$$
Now, to compute $\frac{\partial^2f}{\partial x\partial y}$, you indeed need to apply the product rule, as you have done. For example, the first term would be $$\frac{\partial}{\partial x}\left(\frac{\partial g}{\partial u}(u(x,y),v(x,y))\frac{\partial u}{\partial y}(x,y)\right) = \frac{\partial g}{\partial u}(u(x,y),v(x,y)) \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}(x,y)\right) + \frac{\partial}{\partial x}\left(\frac{\partial g}{\partial u}(u(x,y),v(x,y))\right)\frac{\partial u}{\partial y}(x,y).$$
Since we've been keeping track of where we are computing the derivatives, now you see that in the second term above you have to apply the chain rule again for $$\frac{\partial}{\partial x}\left(\frac{\partial g}{\partial u}(u(x,y),v(x,y))\right).$$
As a sanity check, note that in your computations you end up for instance with $\frac{\partial^2 g}{\partial x \partial u}$. However, $g = g(u,v)$ should be differentiated only with respect to $u$ or $v$, so it hints that something is up with the calculation.