Prove that $\frac{dy}{dx} = -\frac1{(1+x)^2}$ for given that $x\sqrt{1+y} + y\sqrt{1+x} = 0$ $$x\sqrt{1+y} + y\sqrt{1+x} = 0$$

Please tell me where I went wrong. Why I am not getting correct answer ?

There is nothing wrong. Put the value of $y$ to get your result.
However, a simpler approach:
$$x\sqrt{1+y} + y\sqrt{1+x} = 0$$
$$x\sqrt{1+y} = - y\sqrt{1+x}$$
Squarring both sides, we get
$$x^2(1+y) =  y^2(1+x)$$
$$x^2(1+y) -  y^2(1+x)=0$$
$$(x-y)(x+y+xy)=0$$
So either $x-y=0$ or $x+y+xy=0$.
Now if, $x-y=0$, then we have $x=y$. But this does not satisfy the equation $x\sqrt{1+y} + y\sqrt{1+x} = 0$ for all $x,y \in \mathbb{R}$.
Hence required solution is $x+y+xy=0$.
Thus we can write that 
$$1+\frac{dy}{dx}+y+x\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=\frac{-y-1}{1+x}=\frac{\frac{x}{1+x}-1}{1+x}=-\frac{1}{(1+x)^2}$$  
