We have $\dfrac {\partial u }{\partial x} = \dfrac {2x}{x^2 + y^2} $ and $\dfrac {\partial u}{\partial y} = \dfrac {2y}{x^2 + y^2}$

$\therefore \dfrac {\partial ^2 u }{\partial x \partial y} = 2x \left [ - \dfrac {1}{(x^2 + y^2)^2} \right ]2y = \left [ - \dfrac {4xy}{(x^2 + y^2)^2} \right ]$

$ \therefore \dfrac {\partial ^2 u}{\partial y \partial x} = 2y \left [ - \dfrac {1}{(x^2 + y^2)^2} \right ]2x= \left [- \dfrac {4xy}{(x^2 + y^2)^2} \right ]$

$\therefore \dfrac {\partial ^2 u}{\partial x \partial y} = \dfrac {\partial ^2 u}{\partial y \partial x}$