(B)
$\begin{array}{l}
\frac{x}{\sqrt{1+x}}+\frac{y}{\sqrt{1+y}}=0 \\
x \cdot(\sqrt{1+y})+y(\sqrt{1+x})=0 \quad \Rightarrow x \cdot \sqrt{1+y}=-y \sqrt{1+x}
\end{array}$
Squaring both sides we get,
$\begin{array}{l}
x^{2}(1+y)=y^{2}(1+x) \Rightarrow x^{2}+x^{2} y=y^{2}+x y^{2} \\
x^{2}-y^{2}=x y^{2}-x^{2} y \Rightarrow(x-y)(x+y)=-x y(x-y) \\
x+y=-x y \Rightarrow(1+x) y=-x \Rightarrow y=\frac{-x}{1+x}
\end{array}$
Differentiating both sides w.r.t. $x$, we get
$\begin{aligned}
& \frac{d y}{d x}=-\frac{(1+x)(1)-(x)(1)}{(1+x)^{2}}=\frac{-1}{(1+x)^{2}} \\
\therefore &(1+x)^{2} \frac{d y}{d x}=-1
\end{aligned}$