How to take the second derivative using multi variable chain rule?

Question

I am working on this question:

Now I have the first part and found $$\frac{\partial F}{\partial x} = \frac{\partial f}{\partial u}+\frac{\partial f}{\partial v}\\ \frac{\partial F}{\partial y} = \frac{\partial f}{\partial u}-\frac{\partial f}{\partial v}\\ \frac{\partial F}{\partial x}\frac{\partial F}{\partial y} = \left(\frac{\partial f}{\partial u}\right)^2-\left(\frac{\partial f}{\partial v}\right)^2$$ as required.

Yet I am stuck on the second part I know $$\frac{\partial ^2F}{\partial x \partial y}=\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial y}\right)$$ but here is where I am not sure is this:

$$\frac{\partial}{\partial x}\left(\frac{\partial F}{\partial u}-\frac{\partial F}{\partial v}\right)=\frac{\partial F}{\partial u}\frac{\partial u}{\partial x}-\frac{\partial F}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial F}{\partial y}$$

However I think this might be wrong?

Any help?

Answer 1

$$ \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial u}-\frac{\partial f}{\partial v}\right) = \left(\frac{\partial}{\partial u}+\frac{\partial}{\partial v}\right)\frac{\partial f}{\partial u} - \left(\frac{\partial}{\partial u}+\frac{\partial}{\partial v}\right)\frac{\partial f}{\partial v} = \cdots $$