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

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

$x\sqrt{1+y}+y\sqrt{1+x}=0,\;x\ne y.$

$x\sqrt{1+y}=-y\sqrt{1+x}.$

$x^2(1+y)=y^2(1+x).$

$x^2+x^2y=y^2+xy^2.$

$x^2-y^2=xy(y-x).$

$(x-y)(x+y)=-xy(x-y).$

$x+y=-xy.$

$y=-\frac{x}{1+x}.$

$\frac{dy}{dx}=-\frac{1}{(1+x)^2}.$

$\frac{d^2y}{dx^2}=\frac{2}{(1+x)^3}.$

$\displaystyle \frac{d^2y}{dx^2}=\frac{2}{(1+x)^3}.$