I am trying to prove the following
Given a function $f:\Bbb{R}^2\to \Bbb{R}^2$ whose second derivatives are continuous, $$\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}$$
I am aware that proofs for this are found in most books on multivariable calculus. However, I'm trying to come up with a proof on my own.
My strategy is to basically prove that $$\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial x\partial y}$$ should be equal to $0$. What I did was that I performed a change of variables- let $m=x+y$ and $n=x-y$. Then clearly ${x=\frac{m+n}{2}}$ and $y=\frac{m-n}{2}$. I now found the difference $$\frac{\partial^2 f}{\partial m\partial n}-\frac{\partial^2 f}{\partial n\partial m}$$ As $$\frac{\partial f}{\partial m}=\frac{1}{2}\left(\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\right)$$ and $$\frac{\partial f}{\partial n}=\frac{1}{2}\left(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y}\right)$$ we see that $$\frac{\partial^2 f}{\partial m\partial n}-\frac{\partial^2 f}{\partial n\partial m}=\frac{1}{2}\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)$$
My aim would then be this: To prove that $$\frac{\partial^2 f}{\partial m\partial n}-\frac{\partial^2 f}{\partial n\partial m}=\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)$$ Clearly this is only possible if $\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)=0$, as $\frac{1}{2}\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)=\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)$ if and only if $\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)=0$.
Is this a viable strategy? Does anyone know why $$\frac{\partial^2 f}{\partial m\partial n}-\frac{\partial^2 f}{\partial n\partial m}=\left(\frac{\partial^2 f}{\partial x\partial y}-\frac{\partial^2 f}{\partial y\partial x}\right)$$ without assuming the result we are trying to prove?