What is the global maximum of $x^{1/x}$

For $x\gt0$, we have $$ x^{1/x}=e^{\log(x)/x} $$ As $x\to0^+$, $\log(x)/x\to-\infty$.
As $x\to+\infty$, $\log(x)/x\to0$.
In between, $\log(x)/x$ has a maximum of $1/e$ at $x=e$. We get this by looking at its derivative, which is $\frac{1-\log(x)}{x^2}$.

Therefore, for $x\gt0$, we have an infimum of $0$ as $x\to0^+$ and a maximum of $e^{1/e}$ at $x=e$.

For $x\lt0$, let $x=-t$. Then, taking $\log(x)=\log(-t)=\log(t)+i\pi$, we get $$ \begin{align} x^{1/x} &=e^{-(\log(t)+i\pi)/t}\\ &=(\cos(\pi/t)-i\sin(\pi/t))e^{-\log(t)/t} \end{align} $$ As $t\to0^+$ (that is, $x\to0^-$), $-\log(t)/t\to+\infty$.
As $t\to+\infty$ (that is $x\to-\infty$), $-\log(t)/t\to0$.
In between, $-\log(t)/t$ has a minimum of $-1/e$ at $t=e$.

Therefore, for $t>0$, $e^{-\log(t)/t}$ have a minimum of $e^{-1/e}$ at $t=e$ and a supremum of $+\infty$ as $t\to0^+$.

However, since $$ \begin{align} \mathrm{Re}\left(x^{1/x}\right)&=\hphantom{-}\cos(\pi/t)e^{-\log(t)/t}\\ \mathrm{Im}\left(x^{1/x}\right)&=-\sin(\pi/t)e^{-\log(t)/t} \end{align} $$ the fact that $e^{-\log(t)/t}$ grows without bound and the $\sin(\pi/t)$ and $\cos(\pi/t)$ oscillate faster and faster between $+1$ and $-1$ as $t\to0^+$ means that $$ \begin{align} \limsup_{x\to0^-}\mathrm{Re}\left(x^{1/x}\right) &=\limsup_{x\to0^-}\mathrm{Im}\left(x^{1/x}\right)=+\infty\\ \liminf_{x\to0^-}\mathrm{Re}\left(x^{1/x}\right) &=\liminf_{x\to0^-}\mathrm{Im}\left(x^{1/x}\right)=-\infty\\ \end{align} $$ which in turn imply $$ \bbox[5px, border: 1px solid #C00000]{ \begin{align} \sup_{x\lt0}\mathrm{Re}\left(x^{1/x}\right) &=\sup_{x\lt0}\mathrm{Im}\left(x^{1/x}\right)=+\infty\\ \inf_{x\lt0}\mathrm{Re}\left(x^{1/x}\right) &=\inf_{x\lt0}\mathrm{Im}\left(x^{1/x}\right)=-\infty\\ \end{align} } $$

Plots of $\mathrm{Re}\left(x^{1/x}\right)$ and $\mathrm{Im}\left(x^{1/x}\right)$:

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Additional Answers

$\ \small\bullet$ The weird behavior when approaching $0$ from the left is two-fold:

First, for $x\lt0$, $\log(|x|)/x\to+\infty$ as $x\to0^-$; therefore, $\left|\,x^{1/x}\,\right|=e^{\log(|x|)/x}\to+\infty$.

Second, due to the $\pi i$ in $\log(x)$ for $x\lt0$, $\arg\left(x^{1/x}\right)=\pi/x$ forces both real and imaginary parts to oscillate positive and negative.

$\ \small\bullet$ For $x\lt0$, $\mathrm{Re}\left(x^{1/x}\right)^2+\mathrm{Im}\left(x^{1/x}\right)^2=e^{\log(|x|)/x}$ and $\frac{\displaystyle\mathrm{Im}\left(x^{1/x}\right)}{\displaystyle\mathrm{Re}\left(x^{1/x}\right)}=\tan(\pi/x)$.

Yes, of course. Now for those real values.

For x → 0 you have two cases: one is where the function has real values. If you look at x values like -1/n Sr(-1/n) = (-1/n)^-n = ± 1/nⁿ which in absolute value → 0. You'll get the same result with any sequence of x → 0^- for which Sr(x) is real.

Now the case where Sr(x) is complex. Consider for example x = -8/9. (-8/9)^-9/8 = (8/9)^-9/8(-1)^-9/8. The roots of -1 are all on the unit circle, so the absolute value cannot exceed |(-8/9)^-9/8| =(9/8)^9/8, but it could be close if the root you pick has absolute value near 1. Consider |Sr(-4/9)| is near (9/4)^9/4, a larger base and a larger exponent. Since the variance in the absolute value of the roots closest to 1 should be small, it looks as if this kind of sequence will → ∞ You have a few things to nail down to formalize this. Once that is done you can trap other sequences with complex Sr within the sequences you know are → ∞.

However, you've got the problem that some of the subsequences are real and go to zero. So I would say that generally you do not have a limit. If you want the ∞ limit, you are going to have to restrict yourself to sequences where Sr(x) is complex.

You have I think an additional problem, which is that if you pick the zeros of (-1) whose absolute values are closest to 0, the |Sr| may not → ∞.

Hope this helps.

Correct me if im wrong, but is this not Steiner's problem concering the Euler number?

He found it using an inequality, $e^{(x-e)/e} \geq 1 + \dfrac{x-e}{e}$.