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 ?

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, $$y(1+x)=-x\implies y=\dfrac{1-(x+1)}{1+x}=\dfrac1{1+x}-1$$

$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 $

$y = \frac{-x}{x+1} = -1 + \frac{1}{x+1}$

$\frac{dy}{dx} = -(x+1)^2$