Answer
$$ - \frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\sin x + \frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\sin x = 0$$
Work Step by Step
$$\eqalign{
& z = \frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\sin x \cr
& {\text{Find }}\frac{{{\partial ^2}z}}{{\partial {x^2}}}{\text{ and }}\frac{{{\partial ^2}z}}{{\partial {y^2}}} \cr
& \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\sin x} \right] = \frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\cos x \cr
& \frac{{{\partial ^2}z}}{{\partial {x^2}}} = \frac{\partial }{{\partial x}}\left[ {\frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\cos x} \right] = - \frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\sin x \cr
& and \cr
& \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\sin x} \right] = \frac{1}{2}\left( {{e^y} + {e^{ - y}}} \right)\sin x \cr
& \frac{{{\partial ^2}z}}{{\partial {y^2}}} = \frac{\partial }{{\partial y}}\left[ {\frac{1}{2}\left( {{e^y} + {e^{ - y}}} \right)\sin x} \right] = \frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\sin x \cr
& {\text{Substitute into Laplace's equation }}\frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0 \cr
& \underbrace {\frac{{{\partial ^2}z}}{{\partial {x^2}}} + \frac{{{\partial ^2}z}}{{\partial {y^2}}} = 0}_ \downarrow \cr
& - \frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\sin x + \frac{1}{2}\left( {{e^y} - {e^{ - y}}} \right)\sin x = 0 \cr
& 0 = 0 \cr} $$
