$$f(x - f(x/y)) = x f(1 - f(1/y)), \quad \forall x, y \in \mathbb{R}, y \neq 0. \quad (*)$$
Set $y = 1$ in $(*)$ then
$$f(x - f(x)) = x f(1 - f(1)). \tag{1}$$
If $f(t) = 0$ for some $t$ then (1) implies $f(t - f(t)) = f(t) = 0$, hence, in view of
(*), $t = 0$ since $f(1 - f(1)) \ne 0$. On the other hand, (*) gives $f(0 - f(0)) = 0$
so $0 - f(0) = 0$, i.e. $f(0) = 0$. Therefore, $0$ is only zero of $f$.
so $0 - f(0) = 0$, i.e. $f(0) = 0$. Therefore, $0$ is only zero of $f$.
Further, from (1) it follows that $f$ is surjective thus there exist an $\alpha \neq 0$ such
that $f(\alpha) = 1$. Set $y = 1/\alpha$ in (*), then