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.

Well, write $$f(x) = e^{\frac{1}{x}\ln(x)}$$ and differentiate and set equal to 0 to get:
$$\dfrac{d}{dx}(e^{\frac{1}{x}\ln(x)})=\bigg(-\frac{1}{x^2}\ln(x)+\frac{1}{x^2}\bigg)e^{\frac{1}{x}\ln(x)}=0$$
Which implies (after dividing by the exponential term) that
$$\frac{1}{x^2}(1-\ln(x))=0$$
Whence $1=\ln(x)$ or $x=e$.
Now you just need to check whether this gives you a local minimum or maximum via the second derivative.
