values of $x^y=y^x$ for $x \neq y$, hopefully with straightforward explanation As I understand it, many people thought I am asking in a question with exactly this title about pairs of (x,y) such that $x^y=y^x$, or that I am looking for a function $f(x)$, such that $x^{f(x)}=(f(x))^x$, but I am not. I am asking for another function $g(x)=x^{f(x)}$, and where is it concave, convex, its minima, maxima, asymptotes etc. And more importantly, there are two such functions f(x) with the above property and their corresponding functions g(x). Because one $f(x)$ is simply $x$, so for $y=f(x)=x$ we get $g(x)=x^x$, but I am interested in the other function $g(x)$ that is yielded by the rather complicated function $f(x)$.
From the first time I asked the question, I already manage through guesses and observation on individual values come up with these possible properties:
So, with more of a thinking and wolframalpha, I found out (hopefuly this time correctly) that the function is defined only on the interval $(1,\infty)$, goes to infinity for $x$ going from the right to $1$ , it is decreasing  in a convex fashion all the way to $x=e$, where is global minimum $e^e$, and it is increasing towards infinity in rather a quite slow fashion, with a possible asymptote $x=y$, to which it is approaching from the left (from the above), which would indicate it is concave somewhere after $x=e$. Can anybody confirm or correct that?
Is anyone able to help me with this, now that I hope we understand each other?
EDIT: The asymptote part is incorrect, as proved by Ennar. Convexity and concavity are still debatable.
 A: It's correct.
Let $y=kx$, where $k>1$.
Hence,
$$x^y=y^x$$ it's
$$x^{kx}=(kx)^x$$ or $$x=k^{\frac{1}{k-1}}$$ and from here $$y=k^{\frac{k}{k-1}}.$$
Thus, $$x^y=k^\frac{k^{\frac{k}{k-1}}}{k-1}.$$
Let $$f(k)=k^\frac{k^{\frac{k}{k-1}}}{k-1},$$ where $k>1.$ 
Hence,
$$f'(k)=\frac{f(k)\cdot k^\frac{k}{k-1}}{(k-1)^3}\cdot\left(\frac{k-1}{\sqrt k}+\ln{k}\right)\left(\frac{k-1}{\sqrt k}-\ln k\right)>0$$ 
and since $\lim\limits_{k\to1+}f(k)=e^e,$ we obtain $f(k)>e^e$ and we are done!
A: TL;DR You are correct.
So, in the question you were already given the link to, it is stated that $$f(x) = -\frac x{\ln x}W\left(-\frac{\ln x}x\right),$$
however, it is bit imprecise, because $W$ is multivalued function, with its branches denoted with $W_0$ and $W_{-1}$ and thus, if you were to plot $f$ this way in some software that has Lambert W-function implemented, you would get the same graph as by plotting $x^y-y^x=0.$
Thus, the first step would be to eliminate the line $y=x$ to get your $g$.
We will use identities 
\begin{align}
W_0(xe^x) = x,&\ x\geq -1,\\
W_{-1}(xe^x) = x,&\ x\leq -1,
\end{align}
to identify which branch in the above definition of $f$ we need to use:
\begin{align}
f(x) = x &\iff -\frac x{\ln x}W\left(-\frac{\ln x}x\right) = x\\
&\iff W\left(te^t\right) = t,\ x = e^{-t}
\end{align}
By the above identities, we need to use $W_{-1}$ when $t\geq -1$ and $W_{0}$ when $t\leq -1$ to avoid $f(x) = x$. Substituting back with $x$, we get the following definition of $f$:
$$f(x) = \begin{cases}
-\frac x{\ln x}W_{-1}\left(-\frac{\ln x}x\right),\ 0<x< e,\\
-\frac x{\ln x}W_{0}\left(-\frac{\ln x}x\right),\ x\geq e,
\end{cases}$$
but here is the thing, $W_{-1}\left(-\frac{\ln x}x\right)$ is not defined on whole $(0,e)$. The reason is that $W_{-1}$ is only defined on $[-1/e,0)$, so we need to solve $-1/e \leq -\ln x/x < 0$. We get that $x>1$ so the correct definition of $f$ is 
$$f(x) = \begin{cases}
-\frac x{\ln x}W_{-1}\left(-\frac{\ln x}x\right),\ \color{blue}{1<x< e},\\
-\frac x{\ln x}W_{0}\left(-\frac{\ln x}x\right),\ x\geq e.
\end{cases}$$
Finally, we can now give precise definition of $g$ as well (using $x^y = e^{y\ln x}$):
$$g(x) = \begin{cases}
\exp\left(-xW_{-1}\left(-\frac{\ln x}x\right)\right),\ 1<x< e,\\
\exp\left(-xW_{0}\left(-\frac{\ln x}x\right)\right),\ x\geq e.
\end{cases}$$

Let us analize what is happening on $(1,e)$. We want to prove that $-xW_{-1}\left(-\frac{\ln x}x\right)$ is strictly decreasing on $(1,e)$. Let $u(x) = W_{-1}\left(-\frac{\ln x}x\right)$. By differentiating $-xu(x)$ we get $$-u(x) + \frac{(\ln x - 1)u(x)}{(1+u(x))\ln x}$$ and we want to prove it is strictly negative on $(1,e)$. Having in mind that $0<\ln x< 1$ and $u(x)\leq -1$ on $(1,e)$, we can manipulate the expression to get 
$$-u(x) + \frac{(\ln x - 1)u(x)}{(1+u(x))\ln x} < 0\iff u(x)+\frac1{\ln x} > 0.$$
Now, to see this last thing is true involves some trickery which I will skip. It is not as hard as is very tedious. I can write it down if somebody would be interested in that.
We conclude that $-xu(x)$ is strictly decreasing on $(1,e)$ and so is $g(x)$.

Now, the analysis on $[e,+\infty)$ is much easier. If we denote with $v(x) = W_0\left(-\frac{\ln x}{x}\right)$, then $-1 \leq v(x) < 0$ and $v(x)$ is strictly increasing on $(e,+\infty)$ since both $W_0$ and $-\ln x/x$ are. Calculating derivative of $-xv(x)$ we get 
$$-v(x) + \frac{(\ln x - 1)v(x)}{\ln x(1+v(x))}$$
and like the above case, we get that $$-v(x) + \frac{(\ln x - 1)v(x)}{\ln x(1+v(x))}> 0 \iff v(x)\ln x > -1$$ which is true on $(e,+\infty)$ since $v(x)> -1$ and $\ln x> 1$. Thus $-xv(x)$ is strictly increasing on $[e,+\infty)$ and so is $g$.

Having done that, it immediately follows that $g$ has minimum at $x = e$, which is easy to calculate: $g(e) = \exp(-e W_0(-e^{-1})) = \exp(-e\cdot (-1)) = e^e.$
Finally, let us check the limits at boundaries:
$$\lim_{x\to 1^+} g(x) = \exp(\lim_{x\to 1^+} (-xu(x))) = \exp(\lim_{x\to 1^+} (-xW_{-1}(-\frac{\ln x}x))) = +\infty$$ because $\lim_{x\to 0^-}W_{-1}(x) = - \infty.$
To verify that $g(x)\sim x$ at $+\infty$:
$$\lim_{x\to+\infty}\frac{g(x)}{x} = \lim_{x\to+\infty}\frac{\exp(-xv(x))}{x} = \exp(\lim_{x\to+\infty}(-xv(x) - \ln x))$$
and to calculate last limit, we can use Taylor expansion of $W_0$ at $0$:
$$\exp(\lim_{x\to+\infty}(-xv(x) - \ln x)) = \exp(\lim_{x\to+\infty}(-x(-\frac{\ln x}{x}-\frac{\ln^2x}{x^2}+\ldots) - \ln x)) = e^0 = 1.$$
However, $\lim_{x\to +\infty}(g(x)-x) = +\infty$.
