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

$\begin {align*} \text{L.H.S.} &=\sin 2x \dfrac {\partial u}{\partial x}+\sin 2y \dfrac {\partial u}{\partial y} \\ &=\sin 2x \dfrac {\sec^2x}{\tan x + \tan y}+ \sin 2y \dfrac {\sec^2 y}{\tan x+ \tan y} \\ &=\dfrac {1}{\tan x + \tan y}\big[2\sin x \cos x \cdot \sec^2 x + 2 \sin y \cos y \cdot \sec^2 y \big] \\ &= \dfrac {1}{\tan x + \tan y}\big [2\tan x + 2 \tan y \big] \\ &=2\end{align*}$

$=\text{R.H.S}$

Hence, $\sin 2x \dfrac {\partial u}{\partial x} + \sin 2y \dfrac {\partial u}{\partial y} = 2$