In the paper A tribute to Dennis Stanton Richard Askey gives the following problem
If $f(x)=\frac{-8(1-\sqrt{1-x})^3}{x^2}$, then find $f(f(x))$.
I managed to solve it, but spend several hours and used a lot of paper in the process. My proof is not elegant, I just verified it by direct substition. I wonder is there any simple and elegant proof?
Askey writes:
It is a nice problem with a surprising answer, which I leave to the reader. Hint, the first surprise might not be the only one.
So the first surprise is probably the fact that the answer is $x$, which I established.
Question: What is the other surprise that Askey mentions?
If we get rid of the square root in $y=\frac{-8(1-\sqrt{1-x})^3}{x^2}$ by setting $u=(1+\sqrt{1-x})/2$ (those $1+$ and $/2$ are not really needed, but they give nicer formulas), we get $$x=-4(u^2-u),\qquad y=-4(u^{-2}-u^{-1}).$$ So to pass from $x$ to $y$ we replace $u$ by $u^{-1}$, and if we do it again, we get back $u$, and thus the original $x$.
Possibly this is the "official" proof, as the paper that you linked says that Stanton came up with this problem while working with quadratic transformations, and $u\mapsto -4(u^2-u)$ is a quadratic transformation.
And if I'm not supposed to be sloppy: I guess that all the numbers should be real and so $x\leq 1$, and $\sqrt{1-x}\geq0$. For any $x\leq 1$ we have two solutions $u$ of $x=-4(u^2-u)$, namely $(1\pm\sqrt{1-x})/2$; we want the $+$-sign (that's how I defined $u$), i.e. the solution with $u\geq 1/2$. Now if we replace $u$ by $u^{-1}$, we'll keep the inequality $u\geq1/2$ only for $u\in[1/2,2]$. So it's only in this interval that we get $f(f(x))=x$; the interval is $u\in[1/2,2]$, i.e. $x\in[-8,1]$, fitting the graph from @Graviton's comment above.