If $x \sqrt{1+y}+y \sqrt{1+x}=0, x \neq y$ then $2 \frac{d y}{d x}+y$ at x = -3 is equal to

-2

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

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

Squaring both sides

$\Rightarrow x^2(1+y)=y^2(1+x)$

so  $x^2+x^2 y=y^2+y^2 x$

$\Rightarrow x^2-y^2+x y(x-y)=0$

$(x-y)(x+y+x y)=0$

⇒  x + y + xy = 0     (as given  x ≠ y)

at  x = -3

-3 + y - 3y = 0

so  $y = \frac{-3}{2}$

differentiating wrt x

$1+\frac{d y}{d x}+y+\frac{x d y}{d x}=0$

at x = -3 ,   y = -3/2

$1+\frac{d y}{d x}-\frac{3}{2}-3 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-1}{4}$

$\frac{2 d y}{d x}+y=-\frac{1}{2}-\frac{3}{2}=-2$

Option: D