Explanation

(b)
$z = \log{(\tan{x + \tan{y)}}}$
On differentiating partially w.r.t. $x$ and $y$, we get
$\frac{\partial z}{\partial x} = \frac{\sec^{2}x}{\tan{x + \tan y}}\ \ and\ \frac{\partial z}{\partial y} = \frac{\sec^{2}y}{\tan{x + \tan y}}$
Now, $sin\ 2x\ \frac{\partial z}{\partial x} + \sin{2y\frac{\partial z}{\partial y}}$
$= \frac{\sin{2x\sec^{2}{x + \sin{2y\sec^{2}y}}}}{\tan{x + \tan y}}$
$= \frac{2\lbrack\tan{x + \tan{y\rbrack}}}{\tan{x + \tan y}} = 2$