<s>[INST] 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.
 [/INST] the points $(0,0)$ and $(-1,-1)$ no longer lie on the graph.
Btw on actual differentiation it yield  $\frac{dy}{dx}  = \frac{y^2 - 2x(1+y)}{x^2 - 2y(1+x)}$.
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