Question
I'm trying to answer the following question.
By setting $X = \alpha x + \beta y$ and $T = \gamma x + \delta y$, show that the PDE $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial x\partial y} = 0$ can be transformed into $\frac{1}{c^2} \frac{\partial^2 F}{\partial T^2} - \frac{\partial^2 F}{\partial X^2} = 0$, for $f(x,y) = F(\alpha x + \beta y, \gamma x + \delta y)$, where $\alpha, \beta, \gamma, \delta$ are suitably chosen.
Attempt
My first step is to write $\frac{\partial}{\partial x} \left( \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \right) = 0$. My next aim is to write $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ in terms of $ \frac{\partial}{\partial X} $ and $\frac{\partial}{\partial T}$.
I do not understand how to do this. How can you understand how to achieve this in general under a change of variables? Eg, if I have new variables $u(x,y)$ and $v(x,y)$ then in general how can you work out how the $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ are written in terms of the $\frac{\partial}{\partial u}$ and $\frac{\partial}{\partial v}$?
Use the chain rule. You will have, for example, $$ \frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} $$