It's a bit fancy equation .
Let $x>0$ :
Then define the CF in even times:
$$f\left(x\right)=\frac{1}{x+\frac{\cos\left(x\right)}{x+\frac{2\cos\left(\cos\left(x\right)\right)}{x+\frac{3\cos\left(\cos\left(\cos\left(x\right)\right)\right)}{...}}}}$$
Then let one of the maxima over the strictly positive real axis :
$$f'(y)=0$$
Then it seems we have :
$$y=f(y)$$
I have no strategy to tackle it.I discourage the derivative way which is a big problem here.
Question :
Is it true ? Is there a closed form ?
The series does converge, at least for $x > 0$, but we must only take the even terms. I'll use the shorthand $c^n := \cos(...\cos(x)...)$. Writing the continued fractions so all the numerators are 1's, have $f(x) = \frac{1}{x+} \frac{1}{\frac{1}{c^1}x+} \frac{1}{\frac{c^1}{2c^2}x+} \frac{1}{\frac{2c^2}{c^1 3c^3}x +...}$, i.e., $f(x) = [0;\frac{1}{c^1}x, \frac{c^1}{2c^2}x,\frac{2c^2}{c^1 3c^3}x,...].$ From this source https://math.mit.edu/~rmd/IAP/continuedfractions.pdf, one can see that if we only take the even terms, $f_{2n}(x)$, then this gives us an increasing sequence of functions bounded above by $f_1(x)$. Specifically, for each $x$, there is a monotonic sequence $f_{2n}(x)$ which defines a continuous function bounded by $f_1(x)$.
However, the maximum $y$ of $f$ does not seem to satisfy $f(y) = y$. I graphed the first 10 or so $f$'s here: https://www.desmos.com/calculator/jaqljmyshv and you can see that eventually, the $y_{max}$'s lie above $y = x$. We can prove that the $2n+2$th $y_{max}$ always lies above and to the left of the $2n$th $y_{max}$. See that to go from $f_{2n}$ to $f_{2n+2}$ we input $x + \frac{2nc^{2n}}{x+\frac{(2n+1)c^{2n+1}}{x}}$ into the bottom most $x$ slot. When $n >> 0$, of course, $c^n = \cos(...\cos(x)...) \approx 0.7391,$ the dottie number. Simplifying, $x + \frac{2nc^{2n}}{x+\frac{(2n+1)c^{2n+1}}{x}} = \frac{x^3 + (4n+1)cx}{x^2 + (2n+1)c}.$ So for $x \in (0,1)$, as $n$ gets large, this fraction approaches the function $2x$. In general, we go from $f_{2n}$ to $f_{2n+2m}$ by inputting fractions that look like $$x+\frac{nc}{x+\frac{\left(n+1\right)c}{x+\frac{\left(n+2\right)c}{x+\frac{\left(n+3\right)c}{x+\frac{\left(n+4\right)c}{x+\frac{\left(n+5\right)c}{x+...}}}}}}$$ into the bottom most $x$ slot. Via a similar approximation argument, these fractions are close to linear functions with slope $m \in \mathbb{Z}$! Specifically, the fraction with $2k$ layers is $\approx (k+1)x$, assuming $n$ is large. So then your $f$'s can be approximated by $\frac{1}{x + ... } + \frac{1}{mx}$ (assuming you get the indices correct). From here, it is clear that the max of $f_{2n+m +2}$ is attained at a lower $x$ value than $f_{2n+m}$. So since the $f_{max}$'s lie above $y = x$ for the small values I plotted, we know that the limiting case (if it even exists) must lie above $y = x$ as well.