I assume you mean surjective,. Let $f(c)=0$. Take $x=c, a=f(0)$ Take $x=0, f(f(0))=f(0)$ Take $x=f(0), bf(0)^2=0$ If $b=0$, $f(f(x))=a$, and since $f$ is surjective $f$ is constant, contradiction. Hence $f(0)=0$. We thus have $f(f(x))=bxf(x)$ Easy to see $f$ is injective. Can someone help me continue?

parmenides51 wrote: Do there exist numbers $a,b \in R$ and surjective function $f: R \to R$ such that $f(f(x)) = bx f(x) +a$ for all real $x$? Let $P(x)$ be the assertion $f(f(x))=bxf(x)+a$ Since surjective, let $w$ such that $f(w)=0$ $P(w)$ $\implies$ $f(0)=a$ $P(0)$ $\implies$ $f(a)=a$ $P(a)$ $\implies$ $ba^2=0$ $b=0$ would imply $f(f(x))=a$, impossible since surjective, and so $b\ne 0$ and $a=0$ Setting $g(x)=bf(\frac xb)$, surjective, $P(\frac xb)$ becomes New assertion $Q(x)$ : $g(g(x))=xg(x)$ Since surjective, let $u,v$ such that $g(u)=1$ and $g(v)=-1$ (so that $u\ne v$) $Q(u)$ $\implies$ $g(1)=u$ and so $g(g(1))=g(u)=1$ $Q(1)$ $\implies$ $g(g(1))=g(1)=u$ And so $u=1$ $Q(v)$ $\implies$ $g(-1)=-v$ and so $g(g(-1))=g(-v)$ $Q(-1)$ $\implies$ $g(g(-1))=-g(-1)=v$ And so $g(-v)=v$ $Q(-v)$ $\implies$ $-1=-v^2$ and so $v\in\{-1,1\}$ and so $v=-1$ (since $v\ne u=1$) But then $g(-v)=v$ two lines above becomes $g(1)=-1$, in contradiction with previously got $g(1)=1$ And so $\boxed{\text{No such function}}$

Nice solution