Express $\frac{\partial^2F}{\partial x^2} - \frac{\partial^2F}{\partial y^2}$ in terms of the partial derivatives of $F$ with respect to $u$ and $v$. If $F = F(u,v)$ and $u = x - y, v = x + y$, express $\frac{\partial^2F}{\partial x^2} - \frac{\partial^2F}{\partial y^2}$ in terms of the partial derivatives of $F$ with respect to $u$ and $v$. 
I'm not entirely sure how to approach this. Any help would be appreciated. Thank you.

$\frac{\partial u}{\partial x} = 1,\frac{\partial u}{\partial y} = -1, \frac{\partial v}{\partial x} = 1, \frac{\partial v}{\partial y} = 1$
$\frac{\partial F}{\partial x} $
$= \frac{\partial F}{\partial v}\frac{\partial v}{\partial x} + \frac{\partial F}{\partial u}\frac{\partial u}{\partial x}$
$= \frac{\partial F}{\partial v} + \frac{\partial F}{\partial u}$
$\frac{\partial^2 F}{\partial x^2} $
$= \frac{\partial}{\partial v}(\frac{\partial F}{\partial x})\frac{\partial v}{\partial x} + \frac{\partial}{\partial u}(\frac{\partial F}{\partial x})\frac{\partial u}{\partial x}$
$= \frac{\partial}{\partial v}(\frac{\partial F}{\partial x}) + \frac{\partial}{\partial u}(\frac{\partial F}{\partial x})$
$= \frac{\partial}{\partial v}(\frac{\partial F}{\partial v} + \frac{\partial F}{\partial u}) + \frac{\partial}{\partial u}(\frac{\partial F}{\partial v} + \frac{\partial F}{\partial u})$
$=\frac{\partial^2 F}{\partial v^2} +2\frac{\partial^2 F}{\partial v\partial u} + \frac{\partial^2 F}{\partial u^2}$, Assume F is continuous
Similarly,
$\frac{\partial F}{\partial y} $
$= \frac{\partial F}{\partial v}\frac{\partial v}{\partial y} + \frac{\partial F}{\partial u}\frac{\partial u}{\partial y} = \frac{\partial F}{\partial v} - \frac{\partial F}{\partial u}$
$\frac{\partial^2 F}{\partial x^2} $
$= \frac{\partial}{\partial v}(\frac{\partial F}{\partial y}) - \frac{\partial}{\partial u}(\frac{\partial F}{\partial y})$
$=\frac{\partial^2 F}{\partial v^2} -2\frac{\partial^2 F}{\partial v\partial u} + \frac{\partial^2 F}{\partial u^2}$, Assume F is continuous
Therefore,
$\frac{\partial^2 F}{\partial x^2}  - \frac{\partial^2 F}{\partial y^2} = 4\frac{\partial^2 F}{\partial v\partial u}$
