$\dfrac {\partial z}{\partial x} = a \sec^2 (y+ax) - \left ( \dfrac {3a}{2} \right ) (y-ax)^{1/2} $ $\text{and, } \dfrac {\partial^2 z}{\partial x^2} = a^2 2\sec^2 (y+ax)\tan (y+ax) + \left ( \dfrac {3a^2}{4} \right ) (y-ax)^{-1/2} \cdots \ \cdots (1)$ $ \text{now, } \dfrac {\partial z}{\partial y} = \sec^2 (y+ax) + \left ( \dfrac {3}{2} \right )(y-ax)^{1/2}$ $\text{and, }\dfrac {\partial ^2z}{\partial y^2} = 2\sec^2 (y+ax)\tan (y+ax) + \left ( \dfrac {3}{4} \right )(y-ax)^{-1/2} \cdots \ \cdots (2)$ From (1) and (2) $\dfrac {\partial ^2z}{\partial x^2} = a^2 \dfrac {\partial ^2 z}{\partial y^2}$