How to find $f$? I was wondered to solve the following problem:
If $f(y)=f(x)\cdot f(x-y)$ only for $x \neq y$ , what is the $f(x)$ function ? 
thanks
 A: Okay, here's one way to solve the problem. 


*

*First of all, note that either $f$ is the zero function, or it is nonzero everywhere. To prove this, assume that $f(c) = 0$ for some $c$. Then for $x \neq c$, we have $f(x) = f(c) \cdot f(c-x) = 0 \cdot f(c-x) = 0$. Since $f(c) =0$ already by assumption, it follows that $f$ is the zero function. 

*From now on, assume that $f$ is not the zero function. Then claim (1.) implies that $f(x) \neq 0$ for all $x$. 

*For every nonzero $x$, setting $y=x/2$ gives $f(x/2)=f(x)f(x/2)$. Since $f(x/2) \neq 0$, we can cancel it to get $f(x)=1$.

*Finally, $f(0) = f(1) \cdot f(1 - 0) = f(1)^2 = 1$. 
From (3.) and (4.), we can conclude that if $f$ is nonzero, then we must have $f(x) = 1$ for all $x$. So the only solutions are the constant functions $f(x) \equiv 0$ and $f(x) \equiv 1$. 
Depending on what the OP means by "only for $x\ne y$", these two solutions are either the only solutions, or there are no solutions at all.
(We have tacitly assumed that the domain and codomain of $f$ are fields of characteristic $\ne 2$.)
A: Taking $y=0$ we get $f(0)=f(x)^2$ for all $x\ne0$.
If $f(0)=0$ then $f\equiv0$.
If $f(0)\ne0$ then $f$ is never zero.
In this case take $x=2y$ and get $f(y)=f(2y)f(y)$, which implies $f(2y)=1$ for all $y\ne0$. Thus $f\equiv1$.
(Thanks to Henning Makholm and  Srivatsan Narayanan for helping cleaning this up.)
A: I found it.
According to assumption of problem if $y=x$ thus $f(y)\neq f(x).f(x−y)$ or $f(x)\neq f(x).f(0)$, so if $x=0$ thus $f(0)\neq f(0).f(0)=f^2(0)$  since $f(0)\neq 0,1$    $((1))$
If $y=0$ thus $f(0)=f(x)f(x-0)=f^2(x)$ so $f(0)>0$ and if $x \neq0$ thus $f(x)=\pm \sqrt{f(0)} $   $((2))$ 
According to assumption and $((2))$ $\Longrightarrow$  $|f(y)|=|f(x).f(x-y)|$ then $\sqrt{f(0)}=\sqrt{f(0)}.\sqrt{f(0)}=f(0)$ thus $f(0)=0 or 1$ and This is a paradox by using $((1))$, Thus $\forall x,y\in\ D_{f}$ -{0} $\Longrightarrow$ $x-y \notin D_{f}$ $\Longrightarrow$ 
$f(x) =$
\begin{cases}
\ f(0)=c, & \ x=0,
\ 
\pm  \sqrt {c}, & \ x  \ne 0,
\end{cases}
 And $\forall x,y\in\ D_{f}$ -{0} : $x-y \notin D_{f}$ 
Such as $f(x) =$
\begin{cases}
\ 4, & \ x=0,
\ 
-2, & \ x  \in (1,2],
\end{cases}
