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

(A) Given equation: $x \sqrt{1+y} + y \sqrt{1+x} = 0$
Rearranging the terms: $x \sqrt{1+y} = -y \sqrt{1+x}$
Squaring both sides: $x^2(1+y) = y^2(1+x)$
Expanding: $x^2 + x^2y = y^2 + xy^2$
Rearranging: $x^2 - y^2 = xy^2 - x^2y$
Factoring: $(x-y)(x+y) = -xy(x-y)$
Since $x \neq y$ (as $x \sqrt{1+y} = -y \sqrt{1+x}$ implies $x$ and $y$ have opposite signs unless $x=y=0$),we can divide by $(x-y)$:
$x+y = -xy$
$y + xy = -x$
$y(1+x) = -x$
$y = -\frac{x}{1+x}$
Differentiating with respect to $x$ using the quotient rule:
$\frac{dy}{dx} = -\left[ \frac{(1+x)(1) - x(1)}{(1+x)^2} \right]$
$\frac{dy}{dx} = -\left[ \frac{1+x-x}{(1+x)^2} \right] = -\frac{1}{(1+x)^2}$
Hence,proved.