If $x\sqrt{1+y}+y\sqrt{1+x}=0$ and $x\ne y$, then $\frac{dy}{dx}=\_\_\_$

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

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

S.O.B.S.

$x^2\left(1+y\right)=y^2\left(1+x\right)$

$x^2+x^{2}y=y^2+y^{2}x$

$x^2-y^2+x^{2}y-y^{2}x=0$

$\begin{array}{l}\left(x+y\right)\left(x-y\right)+xy\left(x-y\right)=0\\ \left(x-y\right)\left(x+y+xy\right)=0\\ x+y+xy=0\\ x+y\left(1+x\right)=0\\ y\left(1+x\right)=-x\\ y=\frac{-x}{x+1}\\ \frac{dy}{dx}=\frac{-1}{{\left(x+1\right)}^2}\end{array}$