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

The answer I got is $$\frac{dy}{dx}= -\frac{2 \sqrt{1+x} \sqrt{1+y}+y}{x+2 \sqrt{1+x}\sqrt{1+y}}$$ but I cannot simplify it further.
Please provide your assistance.

$$x\sqrt{1+y} = -y\sqrt{1+x}$$
squaring both sides
$$ x^2(1+y) = y^2(1+x)$$
simplifying
$$x^2 - y^2 = xy(y - x)$$
$$x+y = -xy$$
$$y =\frac{-x}{1+x}$$
further take derivatives.
