Find the largest set $A,B$ such that $f:A\rightarrow B, \, \,f(x)=x^x$ has an inverse I want to find the largest interval such that the inverse of the $f(x)=x^x$ function exists.
We have
$$f(x)=x^x\iff \ln (f(x))=(\ln x )e^{\ln x}$$
and $\ln (f(x))=(\ln x) e^{\ln x}$ can be solved iff $\ln (f(x))\ge -\frac 1e\iff f(x)\ge e^{-\frac 1e}≈0.69$.
Obviously, $f(x)=x^x$ doesn't have a real solution for $f(x)<e^{-\frac 1e}$. If, $$-\frac 1e <\ln f(x)<0 \implies e^{-\frac 1e} < f(x)<1$$ then we have an exactly $2$ solution: $W_0$ and $W_{-1}$.
We can not define an inverse function when $e^{-\frac 1e} < f(x)<1\iff x\in(0,1/e)∪(1/e,1)$
My conclusion:

If I define the function $f:\left\{\frac 1e\right\}∪[1,+\infty)\rightarrow \left\{e^{-1/e}\right\}∪[1,+\infty), f(x)=x^x$ then the function $f(x)=x^x$ is invertible and inverse function of $f(x)=x^x$ is $g(x)=e^{W(\ln x)}$.

Am I on the right track?
 A: Yes, but some parts semm to be more complicated than necessary.
TL;DR: The largest interval to define an inverse on is $[1/e^{1/e},\infty)$.
Note: The question has been changed after I wrote the answer below.  The largest interval such that $x^x$ is invertible is $[1/e,\infty)$.
The derivative of $f(x) = x^x$ is
$$f'(x) = \frac d{dx} \left(e^{x\ln x}\right) = (1+\ln x)\,x^x\tag 1$$
For $x\geqslant0$ it's $x^x > 0$, and therefore $f$ is strictly decreasing for $x\in[0,1/e]$ and strictly increasing for $x\in[1/e,\infty)$. The sign of the slope is determined by the factor $1+\ln x$, which is zero iff $x=1/e$ so that $f$ has a global minimum at $(1/e, (1/e)^{1/e})$.
This means we can define inverse functions on the respective domains.  As the Lambert $W$-function satisfies $W(\ln f(x)) = W(x\ln x) =\ln x$ i.e.
$$\exp (W(\ln f(x))) = x\tag 2$$
we get the inverse of $f$ as
$$f^{-1}(x) = \exp(W(\ln x)) = \frac{\ln x}{W(\ln x)}\tag 3$$
More precisely, taking into account the ranges  $W_0\geqslant-1$ and $W_{-1}\leqslant -1$ of the two real branches of $W$:
$W_0 \geqslant -1$
This branch of $W$ leads to the branch
$$\begin{align}
f_0^{-1} : [1/e^{1/e}, \infty) &\to [1/e,\infty) \\
x &\mapsto \exp(W_0(\ln x))
\end{align} \tag{4.1}$$
of the inverse. As $W_0\geqslant -1$, the image is $[1/e,\infty)$. To get a valid argument for $W_0$ we must have $\ln x \geqslant-1/e$ so that the domain of the inverse is $[1/e^{1/e}, \infty)$.
$W_{-1} \leqslant -1$
This branch of $W$ leads to the branch
$$\begin{align}
f_{-1}^{-1} : [1/e^{1/e}, 1] &\to [0, 1/e] \\
x &\mapsto \exp(W_{-1}(\ln x))
\end{align} \tag{4.2}$$
of the inverse.   As $W_{-1}\leqslant -1$, the image is $(0,1/e]$. To get a valid argument for $W_{-1}$ we must have $0 >\ln x \geqslant-1/e$ so that the domain of the inverse is $[1/e^{1/e}, 1)$.
Notice that $\lim\limits_{x\to0+} f(x)=1$ and therefore it is reasonable to have $f^{-1}(1) = 0$.  As $f^{-1}<1/e$, the branch is the one defined by $W_{-1}$, and indeed we get
$$\begin{align}
\lim_{x\to1-}f_{-1}^{-1}(x) 
&= \lim_{x\to1-}\exp (W_{-1}(\ln x)) \\
&= \lim_{x\to0-}\exp (W_{-1}(x)) \\
&= \lim_{x\to-\infty}\exp (x) \\
&= 0 \\
\end{align}$$
so that we can extend that branch to yield $f_{-1}^{-1}(1)=0$.  This is the justification for the definition in (4.2) which includes 1 in the domain (and 0 in the image).
