Partially Correct

$\begin{aligned} u &= \log (\tan x + \tan y) \\ \frac {\partial u}{\partial x} &= \frac {\sec ^2 x}{\tan x + \tan y} \\ \frac {\partial u}{\partial y} &= \frac {\sec ^2 y}{\tan x + \tan y} \end{aligned}$

Therefore, we have:

$\begin{aligned} X &= \sin (2x)\frac {\partial u}{\partial x} + \sin (2y)\frac {\partial u}{\partial y} \\ & = \frac {\sin (2x) \sec ^2 x + \sin (2y) \sec ^2 y}{\tan x + \tan y} \\ & = \frac { \frac {2\sin x \cos x}{\cos ^2 x } +\frac {2\sin y \cos y}{\cos ^2 y }}{\tan x + \tan y} \\ & = \frac {2\tan x + 2 \tan y}{\tan x + \tan y} \\ &= \boxed {2} \end{aligned}$