Baltic Way Shortlist, 2023

Plugging in $y = \frac{a}{x}$ for $a, x \in D$ gives
$$f(f(a)) = 1 - \frac{x}{a f(f(f(x)))}.$$
On the other hand, $y = 1$ and $x = a$ gives for $a \in D$
$$f(f(a)) = 1 - \frac{1}{f(f(f(a)))}.$$
from which we conclude $\frac{x}{f(f(f(x)))}$ is constant for all $x \in D$. Thus it follows $f(f(f(x))) = Cx$ for some constant $C \neq 0$. Plugging this into the second equation gives $f(f(x)) = 1 - \frac{1}{Cx}$. Replace here $x$ with $f(x)$. This is allowed, since $f(x) \in D$ follows from the definition of $f$. Thus
$$Cx = f(f(f(x))) = 1 - \frac{1}{Cf(x)}, \text{ i.e. } f(x) = \frac{1}{C(1 - Cx)}.$$

So we get $f(x) = \frac{1}{1-x}$ for all $x \in D$, which clearly solves the given functional equation, so we are done.