= Solution
The remaining proper <subgroups> are
$$
\boxed{\{1\},\quad\langle g^2\rangle,\quad\langle g\rangle,\quad\langle g^3,g^jh\rangle\ (j=0,1,2),\quad\langle g^2,g^jh\rangle\ (j=0,1).}
$$
Their orders are respectively one, three, six, four and six. The order-four groups are <Klein four-groups>; the last two are <dihedral groups> of order six, isomorphic to $S_3$.
For completeness, if $H$ contains a reflection, write $H\cap\langle g\rangle=\langle g^d\rangle$ with $d\mid6$. Any other reflection differs from a chosen $g^jh$ by an element of this rotational subgroup, so $H=\langle g^d,g^jh\rangle$ with $j$ specified modulo $d$. The choices $d=1,2,3,6$ give respectively the whole group, the two order-six groups, the three order-four groups and the six reflection groups already listed. This proves there are no omitted subgroups.
Thus \b[there are eight remaining proper subgroups]. The printed count nine includes one too many; nine would be correct if the whole group were included with these remaining subgroups. Altogether $G$ has sixteen subgroups, fifteen proper.
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