If $x\sqrt {1 + y} + y\sqrt {1 + x} = 0,$ for $- 1 < x <$1 , prove that $\cfrac{{dy}}{{dx}} = - \cfrac{1}{{{{(1 + x)}^2}}}$

we have, $x\sqrt {1 + y} + y\sqrt {1 + x} = 0$
$\Rightarrow$ $x\sqrt {1 + y} = - y\sqrt {1 + x} \Rightarrow {x^2}(1 + y) = {y^2}(1 + x)$

$\Rightarrow$ $({x^2} - {y^2}) + xy(x - y) = 0 \Rightarrow y = \cfrac{{ - x}}{{x + 1}}$

therefore, $\cfrac{{dy}}{{dx}} = \cfrac{{(x + 1)( - 1) - ( - x)(1)}}{{{{(x + 1)}^2}}} = \cfrac{{ - x - 1 + x}}{{{{(x + 1)}^2}}} = \cfrac{{ - 1}}{{{{(x + 1)}^2}}}$