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