jmc

Setting $x=y=1,$ we get $$f(1)=f(1{)}^2,$$so $f(1)=0$ or $f(1)=1.$

The function $f(x)=0$ shows that $f(1)$ can be 0, and the function $f(x)=x$ (and the function $f(x)=1$) shows that $f(1)$ can be 1. Thus, the possible values of $f(1)$ are $\boxed{0,1}.$