Answer
$$\frac{{\partial z}}{{\partial t}} = - {e^{ - t}}\sin \frac{x}{c} = {c^2}\underbrace {\left( { - \frac{1}{{{c^2}}}{e^{ - t}}\sin \frac{x}{c}} \right)}_{\frac{{{\partial ^2}z}}{{\partial {x^2}}}}$$
Work Step by Step
$$\eqalign{
& z = {e^{ - t}}\sin \frac{x}{c} \cr
& {\text{Find }}\frac{{\partial z}}{{\partial t}}{\text{ and }}\frac{{{\partial ^2}z}}{{\partial {x^2}}} \cr
& \frac{{\partial z}}{{\partial t}} = \frac{\partial }{{\partial t}}\left[ {{e^{ - t}}\sin \frac{x}{c}} \right] = - {e^{ - t}}\sin \frac{x}{c} \cr
& and \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{e^{ - t}}\sin \frac{x}{c}} \right] = {e^{ - t}}\cos \left( {\frac{x}{c}} \right)\left( {\frac{1}{c}} \right) = - \frac{1}{c}{e^{ - t}}\cos \frac{x}{c} \cr
& \frac{{{\partial ^2}z}}{{\partial {x^2}}} = \frac{\partial }{{\partial x}}\left[ { - \frac{1}{c}{e^{ - t}}\cos \frac{x}{c}} \right] = - \frac{1}{{{c^2}}}{e^{ - t}}\sin \frac{x}{c} \cr
& {\text{Therefore,}} \cr
& \frac{{\partial z}}{{\partial t}} = - {e^{ - t}}\sin \frac{x}{c} = {c^2}\underbrace {\left( { - \frac{1}{{{c^2}}}{e^{ - t}}\sin \frac{x}{c}} \right)}_{\frac{{{\partial ^2}z}}{{\partial {x^2}}}} \cr} $$
