Expression of inverse function of $f(x)=\frac{x}{\sqrt{x+1}}$ Here's what I did:
$$f(y)=x$$
$$\frac{y}{\sqrt{y+1}}=x$$
$$\sqrt{y+1}=\frac{y}{x}$$ $$(x\neq0 \,and \,y\neq0)$$
$$y+1=\frac{y²}{x²}$$
$$y²-x²y²-x²=0$$
$$\Delta=x^4+4x²$$
$$y=\frac{x²-\sqrt{x^4+4x²}}{2}\;or\;y=\frac{x²+\sqrt{x^4+4x²}}{2}$$
$y=\frac{x²-\sqrt{x^4+4x²}}{2}$ impossible because $y>-1$
$\\$Then we have:$$f^{-1}(x)=\frac{x²+\sqrt{x^4+4x²}}{2} \forall{x}\in\Bbb{R}$$
But the correct answer seems to be 
$$f^{-1}(x)=\frac{x²+x\sqrt{x^2+4}}{2}$$

Your solution proposes that $f^{-1}(x)$ is always nonnegative, which is not true. 
From $f(x)=\frac{x}{\sqrt{x+1}}$, we can see that $x$ and its image share the same sign. 
Hence from $$y = \frac{x^2\pm \sqrt{x^4+4x^2}}{2}$$
We have \begin{align}y&=\frac{x^2+sign(x)\sqrt{x^4+4x^2}}{2}\\&=\frac{x^2+sign(x)|x|\sqrt{x^2+4}}{2}\\&= \frac{x^2+x\sqrt{x^2+4}}{2}\end{align}
