Obviosly $x=y$ give $f(x)^2=f(0)\to f(0)=0,1$. Because $f(0)=0$ gives $f(x)\equiv 0$, we get $f(0)=1$. Therefore $f(x)=\pm 1$. $y=\frac{x}{2}$ gives $f(x)f(y)=f(y)\to f(x)\equiv 1$.

Can we say: ....$f(x)^2=1$ ($f:\mathbb R\to \mathbb R$)then $f(x)\equiv 1$ or $f(x)\equiv -1$ and the second one is false because $f(0)=1$ Hence $f(x)\equiv 1$ is the only solution?

nicky-glass wrote: Can we say: ....$f(x)^2=1$ ($f:\mathbb R\to \mathbb R$)then $f(x)\equiv 1$ or $f(x)\equiv -1$ and the second one is false because $f(0)=1$ Hence $f(x)\equiv 1$ is the only solution? No, there are more functions satisfying $f(x)^2=1$. An example is $f(x)=1$ for $x\geq0$ and $f(x)=-1$ for $x<0$. You need to do more work to show that $f(x)=1$ is the only solution.

Yes, Thanks.

$P(0,0)\Rightarrow f(0)\in\{0,1\}$ $P(x,x)\Rightarrow f(x)\in\{-1,1\}$ If $f(k)=-1$ then: $P\left(k,\frac k2\right)\Rightarrow f(k)=1$, contradiction, hence $\boxed{f(x)=1}$ which works.

Set $y=x/2$ then $f\equiv 1$ since $f$ not zero. Added. $f\not \equiv 0$ because $f(x)^2 \equiv f(0)$ so $RHS \neq 0.$