Answer
$$\frac{{{\partial ^2}z}}{{\partial {t^2}}} = - 16{c^2}\cos \left( {4x + 4ct} \right) = {c^2}\underbrace {\left[ { - 16\cos \left( {4x + 4ct} \right)} \right]}_{\frac{{{\partial ^2}z}}{{\partial {x^2}}}}$$
Work Step by Step
$$\eqalign{
& z = \cos \left( {4x + 4ct} \right) \cr
& {\text{Find }}\frac{{{\partial ^2}z}}{{\partial {t^2}}}{\text{ and }}\frac{{{\partial ^2}z}}{{\partial {x^2}}} \cr
& \frac{{\partial z}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {\cos \left( {4x + 4ct} \right)} \right] = - 4c\sin \left( {4x + 4ct} \right) \cr
& \frac{{{\partial ^2}z}}{{\partial {t^2}}} = \frac{\partial }{{\partial t}}\left[ { - 4c\sin \left( {4x + 4ct} \right)} \right] = - 16{c^2}\cos \left( {4x + 4ct} \right) \cr
& and \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\cos \left( {4x + 4ct} \right)} \right] = - 4\sin \left( {4x + 4ct} \right) \cr
& \frac{{{\partial ^2}z}}{{\partial {x^2}}} = \frac{\partial }{{\partial x}}\left[ { - 4\sin \left( {4x + 4ct} \right)} \right] = - 16\cos \left( {4x + 4ct} \right) \cr
& {\text{Therefore,}} \cr
& \frac{{{\partial ^2}z}}{{\partial {t^2}}} = - 16{c^2}\cos \left( {4x + 4ct} \right) = {c^2}\underbrace {\left[ { - 16\cos \left( {4x + 4ct} \right)} \right]}_{\frac{{{\partial ^2}z}}{{\partial {x^2}}}} \cr
& \frac{{{\partial ^2}z}}{{\partial {t^2}}} = {c^2}\frac{{{\partial ^2}z}}{{\partial {x^2}}} \cr} $$
