What is the maximum of the self root $f(x) = x^{1/x}$ This is a knowledge sharing question as I have answered it below. I am demonstrating how one would differentiate an expression such as $x^{1/x}$ and proving the following statement.
What is the maximum of the function: $f(x) = x^{1/x}$?
[NOTE] Proof that the found critical point is the maximum is still in progress.
 A: Answer: $e^{e^{-1}}$
Proof:
If we take the derivative of $f(x)$ we can understand where the gradient of the line is zero. The maximum of the curve should the only place where this happens.
Taking the derivative of $f(x)$
$$\frac d {dx}(x^{1/x})$$
Rearrange $x^{1/x}$ as an exponent of $e$
$$\frac d {dx}\left(\exp\left[\frac{\ln x}x\right]\right)$$
Using the chain rule:
$$\frac d {dx}f\circ g = \frac {df}{dg} \frac{dg}{dx}$$
We can rearrange our expression letting $u =\frac {\ln x}x$
$$\frac {d\space\exp(u)}{du} \frac{du}{dx}$$
We can simplify because we are differentiating $e$
$$\exp(u)\frac {du}{dx}$$
Remember $\exp(u) = x^{1/x}$ so now we can write the expression as:
$$x^{1/x}\left( \frac d {dx}\left(\frac {\ln x} x\right) \right)$$
We can now work out the derivative in our expression using the quotient rule:
$$\frac d {dx}\left(\frac uv\right) = \frac {v\frac {du} {dx} - u \frac{dv}{dx}}{x^2}$$
$$\frac d {dx}\left(\frac {\ln x}{x}\right) = \frac {x\frac {d\ln x} {dx} - \ln x\frac{dx}{dx}}{x^2}$$
$$ = \frac {x\frac 1x - \ln x}{x^2} = \frac {1 - \ln x}{x^2} $$
Finally we can now say that
$$\frac d {dx}(x^{1/x}) = x^{1/x}\frac {1 - \ln x}{x^2} $$


Proving that $f(e)$ is the only critical point
There exists a critical point at point $p$ such that $f'(p) = 0$. This implies that there has to be a factor of $0$ in $f'(p)$. As there are only two factors, $\frac {1-\ln x}{x^2}$ has to be the zeroing factor as $x^{1/x} ≠ 0$ this is because no real number $x$ can satisfy $0^x=x$ as $0^n=0$.
For $\frac {1-\ln x}{x^2}$ to equal $0$. The numerator has to be $0$. For $1-\ln x$ to be equal to $0$, $\ln x$ has to be $$\ln x = 1$$ raising both sides as a power of $e$ $$x=e$$
This shows that the only real solution to $f'(p) = 0$ is $e$. This is the only critical point.
Proving that the only critical point is the maximum of $f(x)$
Take a second derivative $f(x) \implies f''(x)$. We can tell if the critical point, which we have identified at $f(e)$ is the maximum or minimum. The concept works as such. Let $k$ be some number such that $g(k)$ is a critical point and $g(x)$ is some function. The following outcomes conclude whether or not our critical is maximum or minimum.
$$g''(k) <0$$Then $g(k)$ is maximum.
$$g''(k) >0$$Then $g(k)$ is minimum.
$$g''(k) = 0$$ Then the test is inconclusive.
$f''(e) < 0 \implies f(e) = \max.$
