Given $x\sqrt{1+y}+y\sqrt{1+x}=0$ $\rightarrow x \sqrt {1+y}=-y\sqrt{1+x}$
Squaring both sides $\rightarrow x^2(1+y)=y^2(1+x)$
$ \Rightarrow x^2 + x^2y-y^2-yx^2=0 \rightarrow (x-y)(x+y+xy)=0$
$ \Rightarrow x+y+xy = 0 \rightarrow y =\large \frac{-x}{1+x}$
Differentiating, we get, $\large\frac{dy}{dx}=\large\frac{-1}{(1+x)^2}$