Differentiate $x \sqrt{1+y}+y \sqrt{1+x}=0$

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.