Answer
$\dfrac{\partial^2 z}{\partial t^2}=a^2 \dfrac{\partial^2 z}{\partial x^2}$
Work Step by Step
Consider $u=x+at; v=x-at$
$\dfrac{\partial z}{\partial x}=(\dfrac{\partial z}{\partial u})(\dfrac{\partial u}{\partial x})+(\dfrac{\partial z}{\partial v})(\dfrac{\partial v}{\partial x})$
Re-arrange as: $\dfrac{\partial z}{\partial x}=f'(u)+g'(v)$
We are given $u=x+at; v=x-at$
and $\dfrac{\partial^2 z}{\partial x^2}=f''(u)(\dfrac{\partial u}{\partial x})+g'(v)(\dfrac{\partial v}{\partial x})$
and $\dfrac{\partial^2 z}{\partial x^2}=f''(u)+g''(v)$
Also, $\dfrac{\partial z}{\partial t}=a f'(u)+g'(v) (-a)=a f'(u)-a g'(v)$
$\dfrac{\partial^2 z}{\partial t^2}=a^2 f''(u)+a^2 g''(v)$
Thus, the result has been proved.
$\dfrac{\partial^2 z}{\partial t^2}=a^2 \dfrac{\partial^2 z}{\partial x^2}$
