<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] $x^2+x^2y = y^2 + xy^2 $
$x^2+x^2y -y^2-xy^2 = 0 $ 
$(x-y)(x+y) + xy(x-y) = 0 $ 
$(x-y)(x+xy+y) = 0 $ 
$(x-y) = 0 , or, x + xy + y = 0 $
$x \neq y$ since $x\sqrt{1+y} = -y\sqrt{1+x}$ unless (0,0)
$x + xy + y = 0 $
$x + y(x + 1) = 0 $
$y = \frac{-x}{x+1} = -1 + \frac{1}{x+1}$
$\frac{dy}{dx} = -(x+1)^2$
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