Explanation

(a)
$\because u = \sin^{- 1}{\left( \frac{x}{y} \right) + \tan^{- 1}\left( \frac{y}{x} \right)}$
$\therefore\ \frac{\partial u}{\partial x} = \frac{1}{\sqrt{\left\{ 1 - \left( \frac{x}{y} \right)^{2} \right\}}}\frac{1}{y} + \frac{1}{1 + \left( \frac{y}{x} \right)^{2}}.\left( - \frac{y}{x^{2}} \right)$
$\Rightarrow x\frac{\partial u}{\partial x} = \frac{x}{\sqrt{(y^{2} - x^{2})}} - \frac{xy}{(x^{2} + y^{2})}\ \ \ \$…(i)
and $\frac{\partial u}{\partial y} = \frac{1}{\sqrt{\left\{ 1 - \left( \frac{x}{y} \right)^{2} \right\}}}\left( - \frac{x}{y^{2}} \right) + \frac{1}{1 + \left( \frac{y}{x} \right)^{2}}.\left( \frac{1}{x} \right)$
$\Rightarrow \ \ y\frac{\partial u}{\partial y} = \ - \frac{x}{\sqrt{(y^{2} - x^{2})}} + \frac{xy}{(x^{2} + y^{2})}$ …(ii)
On adding Eqs. (i)and (ii), we get
$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = 0$