Find all $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy the equation $$f(y+zf(x))=f(y)+xf(z)$$ for all $x,y,z \in \mathbb{R}.$
I can easily see that this function is injective, because of the x without any f in the RHS. I also see that P(x,0,0): $f(0)=xf(0)$, where P(x,y,z) is the assertion, implies f(0)=0. Now P(x,0,1): f(f(x))=xf(1). We also see that if we plug in $x \rightarrow f(x)$, we get that $f(1)=0$, which in turn gives $f(x) \equiv 0$. We also see that if $f(1) \ne 0$, then f is a surjective function. Now I want to prove that f(1)=1 since after that I can continue by induction to prove that f(x)=x, after showing f is multiplicative and additive which is not too hard to prove; it is simple with some simple substitutions. I get a hint that we have $f(r)=1/f(1)$ and then prove that $r= \pm 1$ and then bring a contradiction with $f(1)=-1,$ which is simple enough as $f(-1)=f(f(1))=f(1)=-1$ is a contradiction as $f(1)=-1$ gives an odd function.
Any help will be appreciated!
Since this function is surjective for all non-zero functions $f$, let t be a real number such that $f(t) = 1.$ Substituting $P(t,0,t)$, we get:
$f(t\cdot f(t)) = f(0) + t\cdot f(t)$ $\implies 1 = t$