$\begin{aligned} & x \sqrt{1+y}+y \sqrt{1+x}=0 \Rightarrow x^2(1+y)=y^2(1+x) \\ & \Rightarrow(x-y)(x+y+x y)=0 \Rightarrow x+y+x y=0, \quad\{\because x \neq y\} \\ & \Rightarrow \frac{d y}{d x}=\frac{-1}{(1+x)^2} .\end{aligned}$