Where is the highest point of $f(x)=\sqrt[x]{x}$ in the $x$-axis? I mean, the highest point of the $f(x)=\sqrt[x]{x}$ is when $x=e$.
I'm trying to calculate how can I prove that or how can it be calculated.

It really only makes sense for $x\gt 0$, at least if you stick to real numbers. 
On $(0,\infty)$ can rewrite the function as
$$f(x) = x^{1/x} = e^{(\ln x)/x}.$$
Note that as $x\to\infty$,
$$\lim_{x\to\infty}\frac{\ln x}{x} = 0,$$
so $\lim\limits_{x\to\infty}f(x) = e^0 = 1$ and as $x\to 0^+$, we have
$$\lim_{x\to 0^+}\frac{\ln x}{x} = -\infty$$
so $\lim\limits_{x\to 0^+}f(x) = \lim\limits_{t\to-\infty}e^t = 0$. 
So that means that the function is bounded. We find its critical points by taking the derivative:
$$\begin{align*}
\frac{d}{dx}f(x) &= \frac{d}{dx} e^{(\ln x)/x}\\
&= e^{(\ln x)/x}\left(\frac{d}{dx}\frac{\ln x}{x}\right)\\
&= e^{(\ln x)/x}\left(\frac{x\frac{1}{x} - \ln x}{x^2}\right)\\
&= e^{(\ln x)/x}\left(\frac{1-\ln x}{x^2}\right).
\end{align*}$$
This is zero if and only if $1-\ln x=0$, if and only if $\ln x = 1$, if and only if $x=e$. So the only critical point is at $x=e$.
If $0\lt x \lt e$, then $f'(x)\gt 0$ (since $\ln x \lt 1$), so the function is increasing on $(0,e)$, and if $e\lt x$, then $f'(x)\lt 0$, so the function is decreasing on $(e,\infty)$. Thus, $f$ has a local maximum at $x=e$, and since it is the only local extreme of the function, which is continuous, $f(x)$ has a global extreme at $x=e$.
