Finding the turning points of $f(x)=\left(x-a+\frac1{ax}\right)^a-\left(\frac1x-\frac1a+ax\right)^x$ I've just come across this function when playing with the Desmos graphing calculator and it seems that it has turning points for many values of $a$. 
So I pose the following problem:

Given $a \in \mathbb{R}-\{0\}$, find $x$ such that $\dfrac{dy}{dx}=0$ where $y=\left(x-a+\dfrac1{ax}\right)^a-\left(\dfrac1x-\dfrac1a+ax\right)^x$

As in most maxima/minima problems, we first (implicitly) differentiate it and set to $0$ to give $$\boxed{\small\dfrac{a(ax^2-1)}{x(ax^2-a^2x+1)}\left(\dfrac{ax^2-a^2x+1}{ax}\right)^a=\left(\ln\left(\dfrac{a^2x^2-x+a}{ax}\right)+\dfrac{a(ax^2-1)}{a^2x^2-x+a}\right)\left(\dfrac{a^2x^2-x+a}{ax}\right)^x} \tag{1}$$ I have no idea how to continue from here. I thought about taking logarithms, but it appears to me that the double $\ln$ in the term $\dfrac{a^2x^2-x+a}{ax}$ would only make the equation worse. 
(For the simplest case when $a=1$, the problem is easy: $x=1$ and it is a point of inflexion).

Let's try setting each of the terms to $0$:
Case $1$: $\left(\frac{a^2x^2-x+a}{ax}\right)^x=0$
$ \hspace{1cm}$ This is only possible when the fraction is zero; that is, solving $a^2x^2-x+a=0$ to get $$x=\frac{1\pm\sqrt{1-4a^3}}{2a^2}$$
Case $2$: $\left(\frac{ax^2-a^2x+1}{ax}\right)^a=0$
$ \hspace{1cm}$ This gives $$\begin{align}ax^2-a^2x+1=0&\implies a^2x^2+a=a^3x\\&\implies\left(\dfrac{a^2x^2-x+a}{ax}\right)^x=\left(\dfrac{a^3x-x}{ax}\right)^x=\left(\dfrac{a^3-1}{a}\right)^x=0\end{align}$$ 
$ \hspace{1cm}$ so for equality between LHS and RHS, we must have $a=1$. However, the equation 
$ \hspace{1cm}$ $ax^2-a^2x+1=0$ has no real solutions for such $a$; hence we reach a contradiction.
Case $3$: $\frac{a(ax^2-1)}{x(ax^2-a^2x+1)}=0$
$ \hspace{1cm}$ We have $x=\pm \dfrac1a$. Now LHS is $0$, and $$\left(\dfrac{a^2x^2-x+a}{ax}\right)^x=\left(1-\dfrac1a+a\right)^{\frac1a} \neq 0$$ 
$ \hspace{1cm}$ for $a \in \mathbb{R} - \{\phi\}$, where $\phi$ is the golden ratio. 
$ \hspace{1cm}$ Suppose that $a = \phi$. Then $x$ is forced to be $-\dfrac1a=-\dfrac2{1+\sqrt5}$, since $ax^2-a^2x+1=0$ 
$ \hspace{1cm}$ (undefined) when $x=\dfrac 1a$. This is impossible, since $y$ is only defined when $x>0$ for this 
$ \hspace{1cm}$ value of $a$!
Case $4$: $\ln\left(\frac{a^2x^2-x+a}{ax}\right)+\frac{a(ax^2-1)}{a^2x^2-x+a}=0$
$ \hspace{1cm}$ This is impossible from cases $1$ and $3$. 

UPDATE: I have provided a partial answer to my question, now with $x$ removed from it.

Any hints on how to solve $(3)=(4)$ are welcome.

Here, on MathOverflow: https://mathoverflow.net/questions/302105/on-finding-the-critical-points-of-fx-leftx-a-frac1ax-righta-left-fra

A note about Case $1$ with $\,a:= 2^{-\frac{2}{3}}\,$ and $\,\displaystyle x\to 2^{\frac{1}{3}}\,$ .
Left side $\,=0\,$ for $\,\displaystyle x=2^{\frac{1}{3}}\,$ because of $\, ax^2-1=0\,$  and $\,\displaystyle ax^2-a^2x+1=\frac{3}{2}\ne 0\,$ . 
Right side $\,=0\,$ for $\,\displaystyle x\to 2^{\frac{1}{3}}>1\,$: 
$\,\displaystyle \lim_{z\to +0 \\x>0} z^x\ln z=0\,$ and therefore $\,\displaystyle \left(\frac{a^2x^2-x+a}{ax}\right)^x \ln \frac{a^2x^2-x+a}{ax} \to 0\,$ for $\,\displaystyle x\to 2^{\frac{1}{3}}>0\,$
$\,\displaystyle \left(\frac{a^2x^2-x+a}{ax}\right)^x \frac{a(ax^2-1)}{a^2x^2-x+a} = \left(a^2x^2-x+a\right)^{x-1} (ax^2-1) x^{-x} a^{1-x} = 0\,$  
for  $\,\displaystyle x=2^{\frac{1}{3}}\,$ because of $\,x-1>0\,$ and $\,\displaystyle a^2x^2-x+a=0\,$ and $\, ax^2-1=0\,$ .

If you like to work with recursions for e.g. $\,a>0,\,a\neq 1\,$ 
it can make sense to choose $\,\displaystyle z:=x+\frac{1}{ax}\,$ so that you get   
$\displaystyle x=f_{1,2}(z):=\frac{z}{2}\pm\sqrt{(\frac{z}{2})^2-\frac{1}{a}}\,$ and $\,\displaystyle y=(z-a)^a-\left(az-\frac{1}{a}\right)^x\,$ . 
We get $\,\displaystyle \frac{dy}{dz}=(z-a)^{a-1} - \left(az-\frac{1}{a}\right)^x \left(\frac{x}{z-\frac{1}{a^2}}+\frac{1}{2-\frac{z}{x}}\ln\left(az-\frac{1}{a}\right)\right)$ 
and a possible recursion with $\displaystyle z_0>\max\left(a;\frac{2}{\sqrt{a}};\frac{1}{a^2}\right)\,$ could be 
$\displaystyle z_{n+1}=a+\left(\left(az_n-\frac{1}{a}\right)^{f(z_n)} \left(\frac{f(z_n)}{z_n-\frac{1}{a^2}}+\frac{1}{2-\frac{z_n}{f(z_n)}}\ln\left(az_n-\frac{1}{a}\right)\right)\right)^{\frac{1}{a-1}}\,$ 
with $\,f(z)\in\{f_1(z);f_2(z)\}\,$ .
For time reasons I haven't checked this recursion, sorry. But it's a try to simplify the calculations.   
Note:
Because of $\,\displaystyle \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}\,$ with $\,\displaystyle \frac{dy}{dx}:=0\,$ we have $\,\displaystyle \frac{dy}{dz}=0\,$ or $\,\displaystyle \frac{dz}{dx}=0\,$ .
$\displaystyle \frac{dz}{dx}=0\,$ means $\,ax^2-1=0\,$ which leads directly to my comment about Case $1$ . 
This strengthens the claim that the substitution $\,\displaystyle z:=x+\frac{1}{ax}\,$ makes sense. 
