TITLE: Solving the functional equation $f(xy)=f(f(x)+f(y))$
QUESTION [10 upvotes]: Find all functions from $f: \mathbb{R} \to \mathbb{R}$ such that for all $x$ and $y$  $$f (xy)=f (f (x)+f (y))$$

I've put $x$ and $y$ as $0$ and $1$. How to proceed after substituting if we don't know what the function is?

REPLY [2 votes]: Injective case only:
As suggested by you and in the comments, we use $\{x,y\}=\{0,1\}$:
$y=0$:

$$f(0)=f\big(f(x)+f(0)\big)$$

$x=y=0$:

$$f(0)=f\big(2f(0)\big)$$

so we see: $(1)=(2)$ ->
$$f\big(f(x)+f(0)\big)=f\big(2f(0)\big)\\
f(x)+f(0)=2f(0)\\
f(x)=f(0)=c
$$
We test what $c$ can be: $f(xy)= c = f(2c) = c$ which is true $\forall c \in \mathbb{R}$
So the only solution: $f(x)=c$ with $c \in \mathbb{R}$, (and could easily be generalized to the complex domain $f: \mathbb{C} \to \mathbb{C}$, with $c \in \mathbb{C}$).