$ z = \log(\tan x + \tan y) \\ \text{Partially Differentiating z w.r.t x,}\\ $

$\dfrac{\partial z}{\partial x} = \dfrac{1}{\tan x + \tan y} \times \dfrac{\partial (\tan x + \tan y)}{\partial x} \\ \:\:\: \quad = \dfrac{1}{\tan x + \tan y} \times (\sec^2 x + 0) $

$\\ \therefore, \dfrac{\partial z}{\partial x} = \dfrac{\sec^2 x}{\tan x + \tan y} ...(A)$

$\\ \text{Similarly, Partially Differentiating z w.r.t. y,}\\ \dfrac{\partial z}{\partial y} = \dfrac{\sec^2 y}{\tan x + \tan y} ... (B)$

$\text{Consider, L.H.S }\\ \sin 2x \dfrac {\partial z}{\partial x} + \sin 2y \dfrac {\partial z}{\partial y} = \sin 2x\dfrac{\sec^2 x}{\tan x + \tan y} + \sin 2y \dfrac{\sec^2 y}{\tan x + \tan y} ...\text{From (A) and (B)} $

$ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \: \:= 2\times\sin x \cos x \times \dfrac{1}{\cos^2 x}\times\dfrac{1}{\tan x +\tan y} + 2\sin y \cos y \times\dfrac{1}{\cos^2 y}\times\dfrac{1}{\tan x +\tan y}$

$ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \: \:= 2 \times\dfrac{\sin x }{\cos x}\times\dfrac{1}{\tan x +\tan y} + 2 \times\dfrac{\sin y}{\cos y}\times\dfrac{1}{\tan x +\tan y}$

$ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \: \:= 2 \bigg[\dfrac{\tan x}{\tan x +\tan y} + \dfrac{\tan y}{\tan x +\tan y}\bigg]$

$ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \: \:= 2 \bigg[\dfrac{\tan x + \tan y}{\tan x +\tan y} \bigg] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \: \:=2 = R.H.S. $

Hence,

$\sin 2x \dfrac {\partial z}{\partial x} + \sin 2y \dfrac {\partial z}{\partial y} = 2$