The functional equation and differentiability

Find all functions $f: \mathbb R\rightarrow \mathbb R$, at the same time satisfying the following two conditions:

a) $f (x + yf (x)) = f (x) f (y)$

b) the function $f$ can be represented in the form $f (x) = (\varphi (x)) ^ 2, x \in \mathbb R,$ where the function $f$ has a finite derivative at $x = 0.$ (not infinite)

I have no clue how to start. Any kind of help will be appreciated.

• What do you mean by a 'finite derivative'? Non zero, not infinite? – copper.hat Apr 22 '16 at 16:13
• @ copper.hat: not infinite – Roman83 Apr 22 '16 at 16:15
• $f(0)=f(0)^2$ $f(0)=0$ – kmitov Apr 22 '16 at 16:16
• Is $f(x)=0$ and $f(x)=1$ a solution? Doesn't solve the problem, though. – S.C.B. Apr 22 '16 at 16:18
• @kmitov also $f(0)=1$ is possible – gt6989b Apr 22 '16 at 16:21

Plug in $y=0$ to get $$f(x) = f(x) f(0),$$ so either $f(x) \equiv 0$ or $f(0)=1$.
Now note that $$f(x+yf(x)) = f(y+xf(y))$$ and assuming $f$ is 1-to-1, we have $$x + yf(x) = y + xf(y)\\ x(1-f(y)) = y(1-f(x))\\ \frac{x}{1-f(x)} = \frac{y}{1-f(y)}$$ for arbitrary $x,y$, and that means both LHS and RHS and constant, say $c$.Then you have $$c = \frac{x}{1-f(x)} \\ f(x) = 1 - x/c$$
UPDATE If $f$ is not 1-1, $f(x) \equiv 1$ is a solution, but not sure if there are others...
• Why $$f(x+yf(x)) = f(y+xf(y)) \Rightarrow x + yf(x) = y + xf(y)$$? – Roman83 Apr 22 '16 at 16:38
• @Roman83 if you assume $f$ is one-to-one, then $f(a) = f(b) \implies a = b$. Notice the result satisfies this property as well as that $f(0)=1$... – gt6989b Apr 22 '16 at 16:41