We have $1^{\frac11}=1$ and for any $n>1$, $n^\frac1n > 1$, the minimun and infimum is $1$.
Let $y = x^{\frac1x}$, $$\ln y = \frac{\ln x}{x}$$
$$\frac{d\ln y}{dx}=\frac{d}{dx}\left(\frac{\ln x}{x}\right)= \frac{1-\ln x}{x^2}$$
The value of $y$ increases when $\ln y$ increases. That is when $\frac{d\ln y}{dx}>0$, which is equivalent to $1-\ln x > 0$which is just $\ln x < 1$, taking exponential both sides give us $x < e$.
Hence $y$ increases up to $e$ and then decreases.
Hence for any $x_1, x_2 \in (0,e)$ $x_1 < x_2$ implies that $x_1^{\frac1{x_1}}< x_2^{\frac1{x_2}}$.
For any $x_1, x_2 \in (e, \infty)$ $x_1 < x_2$ implies that $x_1^{\frac1{x_1}}> x_2^{\frac1{x_2}}$.
The only two possible value that could have attained the maximum values are $2$ and $3$.
Since $3^\frac13 > 2^\frac12$, the maximum and supremum is $3^\frac13$.
Remark:
I am working with $\{ n^\frac1n: n \in \mathbb{N} \}$ which is a subset of the real number.
I am not working with $\{ x^\frac1x: x \in \mathbb{R}, x>0 \}$.
Put both derivatives equal to zero: then you have a system
$$
3x^2-63+12y=0 \\
3y^2-63+12x=0
$$
Isolate $y$ in the first equation and you get $y=-\frac{1}{4}x^2+\frac{21}{4}$. Substitute it in the second equation to get $3(-\frac{1}{4}x^2+\frac{21}{4})^2-63+12x=0$ or equivalently $$\frac{3}{16}x^4-\frac{63}{8}x^2+12x-63+3\cdot \frac{441}{16}=0\\
\frac{3}{16}(x^4-42x^2+48x-336+441)=0\\
\frac{3}{16}(x^4-42x^2+64x+105)=0\\
\frac{3}{16}(x-5)(x-3)(x+1)(x+7)=0$$
Now you can find the corresponding $y$'s and the solutions of the systems are $(-7,-7),(-1,5),(3,3),(5,-1)$
You just need to classify them now
Best Answer
$$\tan(nx)=n\tan(x)\iff n\cos(nx)\sin(x)-\sin(nx)\cos(x)$$
Now apply product to sum formulas and double both sides:
$$n\cos(nx)\sin(x)-\sin(nx)\cos(x)=0\iff(1+n)\sin((1-n)x)-(1-n)\sin((n+1)x)=0$$
We have a series solution to:
for the real roots. Here, $\beta=\frac{1+n}{1-n}$, so:
$$\bbox[2px,border:2.5px solid #0D98CA]{\tan(nx)=n\tan(x)\implies x_j=\frac{\pi j}{n+1}+\frac1{2i(n+1)}\sum_{k=1}^\infty\sum_{m=0}^k\frac{(-1)^m}{(k-m)!m!}\left(m+\frac{k-2m}{n+1}-1\right)^{(n-1)}\left(\frac{1+n}{1-n}\right)^k\exp\left(\frac{2\pi i j(m(n-1)+k)}{n+1}\right)}$$
shown here:
using
ReplaceAll[{j->J,n->N}][\[Pi] j/(n+1)+1/(2 I (n+1))Sum[Sum[(-1)^m ((1+n)/(1-n))^k/(m! (k-m)!) Exp[(2 \[Pi] I j ((n-1) m+k))/(n+1)] FactorialPower[m+(k-2 m)/(n+1)-1,k-1],{m,0,k}],{k,1,a}]]for given $a,J,N$. Some cases have closed elementary forms like $n=0,\dots 7$ and for $n=8$, a quintic, has single series or special function solutions. However, for harder cases, the boxed result is a solution. Either sum has a closed form with the Fox H or Fox Wright function too.