<s>[INST] If $x\sqrt{1+y}+y\sqrt{1+x}=0$ find $y'$ 
Find $\frac{dy}{dx}$ if $x\sqrt{1+y}+y\sqrt{1+x}=0$ for $-1\leq x\leq 1$

My Attempt
$$
x\sqrt{1+y}=-y\sqrt{1+x}\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+xy^2\\
2x+2xy+x^2\frac{dy}{dx}=2y\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}\\
\frac{dy}{dx}\Big[ x^2-2y-2xy \Big]=y^2-2x-2xy\\
\frac{dy}{dx}=\frac{y^2-2x-2xy}{x^2-2y-2xy}
$$
How do I proceed further and find the derivative ?
 [/INST] Hint:
Squaring both sides of $$x\sqrt{1+y}=-y\sqrt{1+x}$$ 
$$x^2(1+y)=y^2(1+x)$$
$$\iff0=(x-y)(x+y+xy)$$
But $x\sqrt{1+y}=-y\sqrt{1+x}\implies x,y$ are of opposite sign, hence $x\ne y$ unless $x=y=0$
Otherwise, $$y(1+x)=-x\implies y=\dfrac{1-(x+1)}{1+x}=\dfrac1{1+x}-1$$
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