The only thing you can say that $fg$ has the same smoothness as $g$. That is, if $g$ is in $C^k$, then $fg$ is in $C^k$ as well.
If $f$ in addition has compact support, then $g\in C^k$ implies that $fg$ is in $C^k$ with uniformly continuous derivatives up to order $k$. Similarly, if $g$ is in some Sobolev space $W^{m,p}$, then $fg\in W^{m,p}$ as well.