If $x\sqrt{1+y}+y\sqrt{1+x}=0, x≠ y $ then value of $\frac{d^2y}{dx^2}$ is :

$\frac{2}{(1+x)^3}$ $-\frac{1}{(1+x)^2}$ 0 1

$\frac{2}{(1+x)^3}$