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