Use every non-abelian group of order 6 has a non-normal subgroup of order 2 to classify groups of order 6. 
Prove that every non-abelian group of order $6$ has a non-normal subgroup of order $2$. Use this to classify groups of order $6$.

I proved that every non-abelian group of order 6 has a nonnormal subgroup of order 2, but then how can I use this to classify groups of order 6?

Assuming $H$ is a non normal subgroup of order $2$.
Consider Action of $G$ on set of left cosets of $H$ by left multiplication.
let $\{g_iH :1\leq i\leq 3\}$ be cosets of $H$ in $G$.
(please convince yourself that there will be three distinct cosets)
we now  consider the action $\eta : G\times\{g_iH :1\leq i\leq 3\} \rightarrow \{g_iH :1\leq i\leq 3\}$ by left multiplication.
i.e., take an element $g\in G$ and consider $g.g_iH$ as there are only three distinct sets we get $g.g_iH = g_jH$ for some $j\in \{1,2,3\}$
In this manner, the elements of $g\in G$ takes  cosets $g_iH$ (represented by $i$) to cosets $g_jH$ (represented by $j$) 
i.e., we have map $\eta :\{1,2,3\} \rightarrow \{1,2,3\}$
which can be seen as $\eta : G\rightarrow S_3$
we know that $Ker(\eta)$ is normal in $G$ which is contained in $H$.
As $H$ is not normal in $G$ we end up with the case that $Ker(\eta)=(1)$ i.e., $\eta$ is injective.
i.e., we have $G$ as a subgroup(isomorphic copy) of $S_3$. But, $|G|=|S_3|=6$. Thus, $G\cong S_3$.
So, for any non abelian group $G$ of order $6$ we have $G\cong S_3$.
For an abelian group of order $6$ we have already know that $G$ is cyclic and $G\cong \mathbb{Z}_6$.
So, only non isomorphic groups of order $6$ are $S_3,\mathbb{Z}_6$.
