Differential equation D'Alembert Approach with coordinate transformation. How does one get from 
$$\frac{d^2f}{dz^2}
 - c^2 \frac{d^2f}{dt^2} = 0
$$
with $f $ being $f(z,t)$, by performing a coordinate transformation to get $f(r,s)$
with $r=z-ct$ and $s=z+ct$, to 
$$
\frac{d^2f(r,s)}{dz^2}=\frac{d^2f}{dr^2} +2\frac{d^2f}{drds} + \frac{d^2f}{ds^2}.
$$
and
$$
\frac{d^2f(r,s)}{dt^2}=c^2(\frac{d^2f}{dr^2} -2\frac{d^2f}{drds} + \frac{d^2f}{ds^2}).
$$
 A: Using the chain rule we have:
$$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}$$
From here:
$$\begin{align*}\frac{\partial^2 f}{\partial t^2}&=\frac{\partial}{\partial t}\left(\frac{\partial f}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial f}{\partial s}\frac{\partial s}{\partial t}\right)\\&=\left(\frac{\partial^2 f}{\partial r^2}\frac{\partial r}{\partial t}+\frac{\partial^2 f}{\partial s\partial r}\frac{\partial s}{\partial t}\right)\frac{\partial r}{\partial t}+\frac{\partial f}{\partial r}\frac{\partial^2 r}{\partial t^2}+\left(\frac{\partial^2 f}{\partial r\partial s}\frac{\partial r}{\partial t}+\frac{\partial^2 f}{\partial s^2}\frac{\partial s}{\partial t}\right)\frac{\partial s}{\partial t}+\frac{\partial f}{\partial s}\frac{\partial^2 s}{\partial t^2}\end{align*}$$
Similarly for $\frac{\partial^2 f}{\partial z^2}$.
Now, just plug in the information you have:
$$\frac{\partial r}{\partial t}=-c, \hspace{10pt} \frac{\partial s}{\partial t}=c, \hspace{10pt} \frac{\partial^2 r}{\partial t^2}=\frac{\partial^2 s}{\partial t^2}=0, \hspace{10pt}$$
And remember that if $\frac{\partial^2 f}{\partial r\partial s},\frac{\partial^2 f}{\partial s\partial r}$ are continuous, then $\frac{\partial^2 f}{\partial r\partial s}=\frac{\partial^2 f}{\partial s\partial r}$
