Solving the functional equation $f(x) = f(x / \phi) f(x / \phi^2 - 1)$ I'm trying to find a function $f(x)$ such that the spacing between consecutive roots looks like the infinite Fibonacci word:

$$1, \phi^{-1}, 1, 1, \phi^{-1}, 1, \phi^{-1}, 1, 1, \phi^{-1}, 1, 1, \phi^{-1}, \ldots$$

If I'm not mistaken, any solution to the functional equation $f(x) = f(x / \phi) f(x / \phi^2 - 1)$ must have roots at the points that I want.

And I simply have no idea where to go from here.

How can I solve this functional equation?

Update. I've found that this one similar problem has an easy solution. Change the denominators of $\phi$ and $\phi^2$ both to $2$, so that we have the equation $g(x) = g(x/2) g(x/2 - 1)$. A change of variables gives us this equation:

$$g(2x) = g(x) g(x - 2)$$

Which differs from this double-angle formula only by scaling on the $x$-axis:

$$\sin 2 \theta = \sin \theta \sin (\theta + \pi/2)$$

Thus, we have the easy solution $g(x) = \sin (-\pi x / 4)$. It's not obvious how to apply this solution to the original problem, however.
Substituting $(1+ϕ)x$ for $x$ in your functional equation and transforming according to properties of $ϕ$ leads to the relation $f((1+ϕ)x)=f(ϕx)\cdot f(x-1)$. For $x=1-ϕ$ this implies $f(-ϕ)=f(-1)\cdot f(-ϕ)$, so we must have either $f(-ϕ)=0$ or $f(-1)=1$.
Substituting $(1+ϕ)x$ for $x$ in your functional equation and transforming according to properties of $ϕ$ leads to the relation $f((1+ϕ)x)=f(ϕx)\cdot f(x-1)$. For $x=1-ϕ$ this implies $f(-ϕ)=f(-1)\cdot f(-ϕ)$, so we must have either $f(-ϕ)=0$ or $f(-1)=1$.