Proof using chain rule. For a homework, I need to demonstrate that
$$ \left(\partial z \over \partial x\right)^2 - \left(\partial z \over\partial y\right)^2 = 4{\partial z \over \partial u}{\partial z \over \partial v}$$
where $z = f(x,y)$ and $u = x+y$ and $v = x-y$.
I tried by chain rule, but nothing.
Can anyone help me?
 A: Hint: Consider starting from the right hand side. Use the chain rule. As $x = \frac 12 (u+v)$, $y = \frac 12(u-v)$, you have for example $$ \def\pd#1#2{\frac{\partial #1}{\partial #2}}\pd zu = \pd zx\pd xu + \pd zy \pd yu =\frac 12 \left(\pd zx + \pd zy \right). $$

Addendum. Looking at the question again, I do not know why I wanted to start on the right. Just calculate directly, remembering that for example $$ \pd zx = \pd zu \pd ux + \pd zv \pd vx = \pd zu + \pd zv $$
A: Using the notation $\partial_x$ for $\frac{\partial}{\partial x}$ etc. you have
$$\partial_x u = \partial_y u = \partial_x v = -\partial_y v = 1$$
so now
$$\partial_x z = \partial_u z \partial_x u + \partial_v z \partial_x v = (\partial_u z + \partial_v z)$$
and
$$\partial_y z = \partial_u z \partial_y u + \partial_v z \partial_y v = (\partial_u z - \partial_v z)$$
concluding
$$(\partial_xz)^2 - (\partial_yz)^2 = (\partial_xz + \partial_yz)(\partial_xz - \partial_yz) = 2\partial_uz \cdot 2\partial_vz = 4\partial_uz\partial_vz$$
q.e.d.
