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?

Hint: You know that $G$ has an element of order $3$, this obviously generates a normal subgroup $H$ (why? think index). But, if it then had another subgroup $K$ of order $2$ which is also normal, then $G=HK$, $H\cap K=\{1\}$, and $H,K\unlhd G$. Why is that bad?
For the second part, let $G$ act on something that gives you an injective homomorphism $G\to S_3$. What should it be?
Just for fun, there is a much simpler way to do this. Note that if $H,K$ are above, then $HK=G$, $H\cap K=\{1\}$, and $H\unlhd G$. Thus, $G\cong H\rtimes_\varphi K$. But, note that $\text{Aut}(H)$ has order $2$, and so there are at most two distinct groups of order $6$. Since $\mathbb{Z}/6\mathbb{Z}$ and $S_3$ are distinct, they must constitute all of them. This acdtually shows you how to classify groups of order $2p$, $p$ an odd prime.
