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.

Answer

Here is a possible approach.

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...