What is the kernel of $h$ where $h:S_3 to mathbb{R}^*$

If $h: S_3 to R^*$ is a homomorphism, then $S_3/textrm{kern}(h) cong h(S^3) subseteq R^*$. For $sigma, tau in S^3$, we would then have $$h(sigma tau sigma^{-1} tau^{-1}) = h(sigma) h(tau) h(sigma^{-1}) h( tau^{-1}) = h(sigma) h(tau) h(sigma)^{-1} h( tau)^{-1} = 1$$ by the commutativity of $R^*$. Therefore $textrm{kern(h)}$ contains the commutator subgroup of $S_3$. This subgroup happens to be the cyclic group generated by $(1;2;3)$. By lagrange's theorem, since the order of $textrm{kern(h)}$ is at least 3, we have that $|h(S_3)| leq 2$ which leaves few options for $h$.

The imagine of $S_3$ must be a subgroup of $Bbb R ^*$ of order a divisor of 6. So, if $xin h(S_3)$ $x^6=1implies x=pm 1$. Now we have two possibilities: $h(S_3)={1}$ or $h(S_3)={pm 1}$. In the first case the kernel is all the group, in the second it must be a subgroup of $S_3$ of index 2, so it must be $A_3$.