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