Let $z = \phi(u,v),u = f(x,y), v = g(x,y)$ with $\phi, f, g$ infinitely smooth. Suppose $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$. Prove the following equalities:
$$ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} $$
and $$ \frac{\partial^2 \hat{\phi}}{\partial x^2} + \frac{\partial^2 \hat{\phi}}{\partial y^2} = \left[ \left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial v}{\partial x}\right)^2 \right] \left[ \frac{\partial^2 \phi}{\partial u^2} + \frac{\partial^2 \phi}{\partial v^2} \right] $$
with $\hat{\phi} = \phi(f(x,y), g(x,y))$.
I think I managed to prove the first equality, I just worked with each side of the equality and used $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$. This, together with the Schwarz Theorem for change the order of derivation, allowed me to conclude that both sides of the first equality are $0$, and in particular, are equal.
However, for the second equality, I am not quite sure about what to do. I know from the Chain Rule that $\frac{\partial \hat{\phi}}{\partial x} = \frac{\partial \phi}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial \phi}{\partial v}\frac{\partial v}{\partial x}$, and the same for variable $y$. But I don't know how to continue. The exercises I have done previously involved concrete functions, maybe I just have a bad day, but I'm feeling lost. Can you help?
I finally managed it! The key was noting that $\frac{\partial}{\partial x}\left( \frac{\partial \phi}{\partial u}\right) = \frac{\partial^2 \phi}{\partial u^2} \frac{\partial u}{\partial x} + \frac{\partial^2 \phi}{\partial u\partial v}\frac{\partial v}{\partial x}$. This allow us to compute the second partial derivatives of $\phi$ using the product rule. The rest is just calculations using the first equality for simplifying terms and the hypothesis at the very end.