Explanation

(b)
Given, $x\sqrt{1 + y} = - y\sqrt{1 + x}\$ …(i)
On squaring both sides, we get
$x^{2}(1 + y) = y^{2}(1 + x)$
$\Rightarrow$ ($x - y)(x + y) + xy(x - y) = 0$
$\Rightarrow \ (x - y)(x + y + xy) = 0$
$x - y\ \neq 0\ \ because\ it\ does\ not\ satisfy\ the\ Eq.(i)$.
$\therefore\ \ x + y + xy = 0\ \ \Rightarrow \ \ y = - \frac{x}{1 + x}$
$\Rightarrow \ \frac{dy}{dx} = - \frac{(1 + x)(1) - x(1)}{(1 + x)^{2}} = - \frac{1}{(1 + x)^{2}}$