Answer
$${\frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0}$$
Work Step by Step
$$\eqalign{
& z = 5xy \cr
& {\text{Find }}\frac{{{\partial ^2}z}}{{\partial {x^2}}}{\text{ and }}\frac{{{\partial ^2}z}}{{\partial {y^2}}} \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {5xy} \right] = 2y \cr
& \frac{{{\partial ^2}z}}{{\partial {x^2}}} = \frac{\partial }{{\partial x}}\left[ {2y} \right] = 0 \cr
& and \cr
& \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {5xy} \right] = 5x \cr
& \frac{{{\partial ^2}z}}{{\partial {y^2}}} = \frac{\partial }{{\partial y}}\left[ {5x} \right] = 0 \cr
& {\text{Substitute into Laplace's equation }}\frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0 \cr
& \underbrace {\frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0}_ \downarrow \cr
& 0 + 0 = 0 \cr
& 0 = 0 \cr} $$
