How to we get the inverse function? Let $f(x)=\sqrt{\frac{e^x-1}{x}}, x \neq 0$, $f(0)=1$.
It is given that $g=f^2$.
I want to show that the function $f^{-1}:(0,+\infty) \to \mathbb{R}$ is a function and that the equation $f(x)=f^{-1}(x)$ has at least 2 solutions, $x_2>x_1>1$.
Also, show that $x_1, x_2$ are the only solutions of the above equation and that there is a unique $\xi>1$ such that $g(\xi)+2\xi f(\xi)=e^{\xi}$. (we suppose that $f$ is convex)
According to wolframalpha, the inverse function is the following:

What exactly is $W$ ? And how to we come to the inverse of the function?
EDIT: I have thought the following:
WE have that $f(f^{-1}(x))=x$. Thus we get that $\frac{e^{f^{-1}(x)-1}}{f^{-1}(x)}=x^2$.
Does this help ? Can we find like that the formula of the function $f^{-1}$ ?
 A: Lambert's $W$ is defined as the function $W(x)$ that satisfies the relationship
$$W(x)e^{W(x)}=x\tag1$$
Let $y(x)=f^{-1}(x)$ denote the inverse function of the function
$$f(x)=\begin{cases}\sqrt{\frac{e^x-1}{x}}&,x\ne 0\\\\1&,x=0\end{cases}\tag2$$
Then it is easy to see from $(2)$ that $y(x)$ satisfies the equation
$$e^{y(x)}=x^2\left(y(x)+\frac1{x^2}\right)\tag3$$
Substituting $s(x)=-\left(y(x)+\frac1{x^2}\right)$ into $(3)$ yields
$$s(x)e^{s(x)}=-\frac{e^{-1/x^2}}{x^2}\tag4$$
Comparing $(4)$ with $(1)$ reveals
$$s(x)=W\left(-\frac{e^{-1/x^2}}{x^2}\right)$$
whereupon we find that
$$f^{-1}(x)=-W\left(-\frac{e^{-1/x^2}}{x^2}\right)-\frac1{x^2}$$
which agrees with the result from Wolframalpha that the OP reported.


Now, let's take a step backward.  First, the function $f(x)$ as given in $(2)$ is differentiable with $f'(x)>0$ on $[0,\infty)$.  Therefore, $f(x)$ has a unique inverse function for $x\ge 0$.

Next, let's show that $f(x)=x$ has two roots. Note that $f(x)>x$ for $x=0$.
Then, using $e<3$ it is easy to see that $f(2)=\sqrt{\frac{e^2-1}{2}}<2$.  So, $f(x)<x$ at $x=2$.
Finally, since $e^x-1\ge \frac{x^4}{24}$, we find that $f(x)\ge \frac{x^{3/2}}{24}$ and hence $f(x)>x$ for $x>24$.
Putting this together, we see that $f(x)=x$ for two distinct values ox $x$, $x_1$ and $x_2$ such that $0<x_1<2$ and $2< x_2<24$.  A numerical root search finds that $x_1   \approx 1.545$ and $x_2\approx 4.567$.

Note that at any fixed point $x_0$ of $f$, $f(x_0)=x_0\implies x_0=f^{-1}(x_0)$.  Therefore, $x_1$ and $x_2$ are also solutions of the equation $f(x)=f^{-1}(x)$.
A: $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
The wolframalpha solution
is incomplete.
It should be
\begin{align}
f^{-1}(x)
&=
\begin{cases}
f_0^{-1}(x)=
-\Wp(-\tfrac1{x^2}\exp(-\tfrac1{x^2}))- \tfrac1{x^2}
,\quad & x\in(0,1)
,\\
f_1^{-1}(x)=
-\Wm(-\tfrac1{x^2}\exp(-\tfrac1{x^2}))- \tfrac1{x^2}
,\quad & x\in[1,\infty)
,
\end{cases} 
\end{align}
where $\Wp$ is the principal branch
and $\Wm$ is the other real branch
of the Lambert $\W$ function.
And indeed the equation $f(x)=f^{-1}(x)$
has two real solutions, which can be fouund numerically as
\begin{align}
x_1&\approx 1.5450
,\\
x_2&\approx 4.5670
.
\end{align}
$\endgroup$

