Selection and Training Session

$$f(x-f(x/y))=xf(1-f(1/y)),\forall x,y\in \mathbb{R},y\ne 0.(\ast )$$ Set $y=1$ in $(\ast )$ then $$f(x-f(x))=xf(1-f(1)).\text{\hspace{1em}(1)}$$ 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 $\alpha \ne 0$ such that $f(\alpha )=1$. Set $y=1/\alpha$ in (*), then