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

This is not an answer, just a comment.
Your $f(a,x)=\left(x-a+\dfrac1{ax}\right)^a-\left(\dfrac1x-\dfrac1a+ax\right)^x$ gives us for different $a$ functions of $x$ that probably have different domains.
So, when you say "Given $a \in \mathbb{R}$" that is not at all as precise as it could be. It would be more precise if we could for every $a\in \mathbb R$ calculate the domain of $f(a,x)$, because, when you differentiated this function in order to find critical point(s), are you sure that they are in the domain of $f$, even if you succeed in calculating them?
It is natural to think that as $a$ vary continuously that then the domains of $f$ will vary continuously but it is a question of what happens for $a=0$, that is, does $\dfrac {1}{f(a,x)}$ has limit when $a \to 0$, or is discontinuous (in the sense of domains)?
I am of the opinion that finding solutions to your (1) should not be done before first calculating for which $a \in \mathbb R$ your $f$ is well-defined function of $x$, and what is the domain of $f$ for every $a \in \mathbb R$.
