I want to find the local extrema of $x^{\frac{x}{x!}}$ for all $x>0$. Using Desmos, I got $(0.379,0.661)$ and $(2.228,2.046)$. What I want to do is find the exact values.
Here is what I tried: $$f(x)=x^{\frac{x}{x!}}$$
$$f'(x)=\frac{x^{\frac{x}{x!}}\left(x!\left(1+\ln\left(x\right)\right)-x\Gamma\left(1+x\right)\ln\left(x\right)\psi\left(1+x\right)\right)}{{x!}^{2}}$$
$$0=\frac{x^{\frac{x}{x!}}\left(x!\left(1+\ln\left(x\right)\right)-x\Gamma\left(1+x\right)\ln\left(x\right)\psi\left(1+x\right)\right)}{{x!}^{2}}$$
$$0=x^{\frac{x}{x!}}\left(x!\left(1+\ln\left(x\right)\right)-x\Gamma\left(1+x\right)\ln\left(x\right)\psi\left(1+x\right)\right)$$
$$0=x!(1+\ln(x))-x\Gamma(1+x)\ln(x)\psi(1+x)$$
$$0=1+\ln(x)(1-x\psi(1+x))$$
$$0=1+\ln(x)(1-x(\psi(x)+\frac{1}{x}))$$
$$0=1+\ln(x)(1-x\psi(x)-1)$$
$$0=1+\ln(x)(-x\psi(x))$$
$$0=\ln(e)-\ln(x^{x\psi(x)})$$
$$0=\ln(\frac{e}{x^{x\psi(x)}})$$
$$e^0=\frac{e}{x^{x\psi(x)}}$$
$$1=\frac{e}{x^{x\psi(x)}}$$
$$e=x^{x\psi(x)}$$
I don't know how to solve it from here.
After resolving the problem with the suggestions in the comments (I appreciate them), I came to the same conclusion.
As @Thomas Andrews commented, it is simpler to find the extrema of function $$g(x)=\log(f(x))=\frac {\log(x)}{\Gamma(x)}$$ Since $$g'(x)=\frac{1-x \log (x) \,\psi ^{(0)}(x)}{x \Gamma (x)}$$ the problem is to find the zeros of function $$h(x)=1-x \log (x) \,\psi ^{(0)}(x)$$ This is what you wrote and, as you concluded there is (more than likely) no explicit solution even with speacial functions.
However, without resorting to numerical methods, we can obtain the solution using series expansions.
For the smallest root, expand $h(x)$ as a series around $x=\frac 1 e$ $$h(x)=\sum_{n=0}^\infty \frac{a_n}{e\, n!}\,\left(x-\frac{1}{e}\right)^n$$ where all $a_n$ express in terms of successive polygamma functions.
$$\left( \begin{array}{cc} n & a_n \\ 0 & e+\psi ^{(0)}\left(\frac{1}{e}\right) \\ 1 & \psi ^{(1)}\left(\frac{1}{e}\right) \\ 2 & \psi ^{(2)}\left(\frac{1}{e}\right)-e^2 \psi ^{(0)}\left(\frac{1}{e}\right) \\ 3 & e^3 \psi ^{(0)}\left(\frac{1}{e}\right)-3 e^2 \psi ^{(1)}\left(\frac{1}{e}\right)+\psi ^{(3)}\left(\frac{1}{e}\right) \\ 4 & -2 e^4 \psi ^{(0)}\left(\frac{1}{e}\right)+4 e^3 \psi ^{(1)}\left(\frac{1}{e}\right)-6 e^2 \psi ^{(2)}\left(\frac{1}{e}\right)+\psi ^{(4)}\left(\frac{1}{e}\right) \\ \end{array} \right)$$
Now, using power series reversion, we can write $$x=\sum_{n=0}^\infty b_n\, \left(h(x)-h\left(\frac{1}{e}\right)\right)^n$$ and remember that we want $h(x)=0$. Since we know all the $a_n$, using Morse and Feshbach method, we know all the $b_n$.
So, we have the exact result as the result of an infinite summation which is the case for many mathematical functions.
Just to give an idea, limiting the expansion to the only terms given in the above table, converted to decimals, the result is $$x=\color{red}{0.379261164}06$$.
Still for illustration, let us add more and more terms in the truncated initial series expansion $$\left( \begin{array}{cc} n & x_{(n)} \\ 4 & 0.379261164057556202762 \\ 5 & 0.379261164944292392549 \\ 6 & 0.379261164964225130129 \\ 7 & 0.379261164964723006970 \\ 8 & 0.379261164964736066201 \\ 9 & 0.379261164964736408755\\ 10 & 0.379261164964736417981 \\ 11 & 0.379261164964736418234 \\ 12 & 0.379261164964736418242 \\ \end{array} \right)$$
We can do the same for the largest root writing $$h(x)=\sum_{n=0}^\infty \frac{a_n}{e^{n-1}\, n!}\,\left(x-e\right)^n$$ where all $a_n$ express again in terms of successive polygamma functions. $$\left( \begin{array}{cc} n & a_n \\ 0 & \frac{1}{e}-\psi ^{(0)}(e) \\ 1 & -2 \psi ^{(0)}(e)-e \psi ^{(1)}(e) \\ 2 & -\psi ^{(0)}(e)-4 e \psi ^{(1)}(e)-e^2 \psi ^{(2)}(e) \\ 3 & \psi ^{(0)}(e)-3 e \psi ^{(1)}(e)-6 e^2 \psi ^{(2)}(e)-e^3 \psi ^{(3)}(e) \\ 4 & -2 \psi ^{(0)}(e)+4 e \psi ^{(1)}(e)-6 e^2 \psi ^{(2)}(e)-8 e^3 \psi ^{(3)}(e)-e^4 \psi ^{(4)}(e) \\ \end{array} \right)$$
It will be less spectacular because the starting point is a bit too far from the solution $$\left( \begin{array}{cc} n & x_n \\ 4 & 2.23064895313597364924 \\ 5 & 2.22910675318805202320 \\ 6 & 2.22849746073697814456 \\ 7 & 2.22824517539565051668 \\ 8 & 2.22813712515341018260 \\ 9 & 2.22808965433751758028 \\ 10 & 2.22806837892986579765 \\ 11 & 2.22805868996394215244 \\ 12 & 2.22805421920404357243 \\ \end{array} \right)$$
Continuing the processes $$x_1=0.3792611649647364182419270048931725660452143622594846919\cdots$$ $$x_2=2.2280502540931513533282382617824801289534669840038862232\cdots$$
These numbers are not recognized by the inverse symbolic calculators I have been able to run.