If I have a function $f(t-kx)$ where $k$ is a constant and $t$ and $x$ are variables, then I know by looking at it that $\frac{\partial f}{\partial t}=-(1/k)\frac{\partial f}{\partial x}$ since it's obvious that changing $x$ is equivalent to changing $t$.
How do I show this formally? What is the right terminology for this type of problem and solution, and can it be generalized to more complicated cases like $f=f(u(t,x))$ for some function $u$?
Alternatively using $u=t-kx$ I can try $\frac{\partial f}{\partial t}=\frac{\partial f}{\partial u} \frac{\partial u}{\partial t}$, and since $\frac{\partial u}{\partial t}=1$ this gives $\frac{\partial f}{\partial t}=\frac{\partial f}{\partial u}$ which is not helpful, but if I apply the chain rule twice then $\frac{\partial f}{\partial t}=\frac{\partial f}{\partial u} \frac{\partial u}{\partial t}=\frac{\partial f}{\partial x} \frac{\partial x}{\partial u} \frac{\partial u}{\partial t}$, for the simple case $u=t-kx$, I can find $\frac{\partial x}{\partial u}=-(1/k)$ but it seems this is only possible if the function $u(t,x)$ is invertible for $x$?
Let $$u=t-kx$$
$$\frac {\partial {f}}{\partial t}=\frac {\partial {f}}{\partial u}.\frac {\partial u}{\partial t} =\frac {\partial {f}}{\partial u}$$ Similarly you find $$\frac {\partial f}{\partial x}=\frac {\partial {f}}{\partial u}.\frac {\partial u}{\partial x}=-k\frac {\partial {f}}{\partial u}$$
You can take over from here.