Explanation

(c)
On differentiating partially the given equation w.r.t. $x$ and $y$
$\frac{\partial u}{\partial x} = \frac{(x + y)}{(x^{2} + \ y^{2})} \times \frac{x^{2} - y^{2} + 2xy}{(x + y)^{2}}$
$\Rightarrow \ x\frac{\partial u}{\partial x} = \frac{x}{(x^{2} + y^{2})} \times \frac{x^{2} - y^{2} + 2xy}{(x + y)}$ …(i)
and $\frac{\partial u}{\partial y} = \frac{(x + y)}{(x^{2} + y^{2})} \times \frac{y^{2} - x^{2} + 2xy}{(x + y)^{2}}$
$\Rightarrow y\frac{\partial u}{\partial y} = \frac{y}{(x^{2} + y^{2})} \times \frac{(y^{2} - x^{2} + 2xy)}{(x + y)}$ …(ii)
On adding Eqs. (i) and (ii), we get
$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{1}{\left( x^{2} + y^{2} \right)(x + y)}$
$\lbrack x\left( x^{2} - y^{2} + 2xy \right) + y\left( y^{2} - x^{2} + 2xy \right)\rbrack$
$= \frac{1}{\left( x^{2} + y^{2} \right)(x + y)} \times \left( x^{2} + y^{2} \right)(x + y)$
$= 1$