Answer
$\dfrac{\partial^2 z}{\partial t^2}=a^2 \dfrac{\partial^2 z}{\partial x^2}$
Work Step by Step
$\dfrac{\partial z}{\partial x}=(\dfrac{\partial z}{\partial u})(\dfrac{\partial u}{\partial x})+(\dfrac{\partial z}{\partial v})(\dfrac{\partial v}{\partial x})=f'(u)+g'(v)$
and $z_{xx}=f''(u)(\dfrac{\partial u}{\partial x})+g'(v)(\dfrac{\partial v}{\partial x}) \\ =f''(u)+g''(v)$
$z_t=(\dfrac{\partial z}{\partial u})(\dfrac{\partial u}{\partial t})+(\dfrac{\partial z}{\partial v})(\dfrac{\partial v}{\partial t}) \\=f'(u) \cdot a+g'(v) (-a) \\=a f'(u)-a g'(v)$
Now, $z_{tt}=a^2 f''(u)+a^2 g''(v)$
So, $\dfrac{\partial^2 z}{\partial t^2}=a^2 \dfrac{\partial^2 z}{\partial x^2}$
