$\frac{\partial^2 u}{\partial t^2}-c^2\frac{\partial^2u}{\partial x^2}=0\text{ implies } \frac{\partial^2 u}{\partial z \partial y}=0$

Compute the $(x,t)$ derivatives in terms of the new variables $(z,y)$ by means of your given transformation. Then, the chain rule reads:

\begin{align} \frac{\partial u}{\partial t} = & \frac{\partial u}{\partial z } \frac{\partial z }{\partial t } + \frac{\partial u}{\partial y } \frac{\partial y }{\partial t } = c \, (u_z - u_y) \\ \frac{\partial u}{\partial x} = & \frac{\partial u}{\partial z } \frac{\partial z }{\partial x } + \frac{\partial u}{\partial y } \frac{\partial y }{\partial x } = u_z + u_y \\ \frac{\partial^2 u}{\partial t^2} = & \frac{\partial}{\partial t}\left( c \, (u_z - u_y) \right) = c^2 (u_{zz} -2 u_{zy} + u_{yy}) \end{align}

Can you derive the second $x$-derivative and substitute back in the original PDE?

Hope this helps. Cheers!

Write the variables $x$ and $t$ in terms of $z$ and $y$. Next use the chain rule by writing

$$u(x,t) = u(\phi(z,y), \psi(z,y)).$$

Next compute

$$\frac{\partial}{\partial y} u(\phi(z,y), \psi(z,y))$$

and

$$\frac{\partial}{\partial z} \left( \frac{\partial}{\partial y} u(\phi(z,y), \psi(z,y)) \right).$$

You will use the equivalent form of the wave equation given by

$$\frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} =0.$$