For a proof by contradiction, you assume the truth of the premise and the negation of the desired conclusion to obtain a contradiction:
"$3n+2$ is odd then $n$ is odd"
Assume both
(1) $3n + 2$ IS odd, and hence, there $3n + 2 = 2k+1$ for some integer $k$.
(2) $n$ is even, that is, there is some integer $m$ such that $n = 2m$.
From these assumptions, you obtain a contradiction, from which you conclude that the negation of both $(1)$ and $(2)$ entails the affirmation of the statement to be proved:
That it must follow that if $3n+2$ is odd, then $n$ is odd.
In general, a proof by contradiction proceeds as follows:
Prove: $\;p\rightarrow q$.
Assume $\;\lnot (p \rightarrow q) \equiv \lnot(\lnot p \lor q) \equiv (p \land \lnot q)$.
Derive a contradiction.
Conclude $\;\lnot \lnot (p \rightarrow q) \equiv (p\rightarrow q)$, as desired.
Added:
For a general discussion about how the differences and similarities between the two proof strategies, you might want to read an earlier MSE post:
Best Answer
You did very well: you got to the "meat" of the proof.
I'll simply add "a side dish":
I would simply add, after demonstrating that, given $m$ and $n$ are both odd, and hence, as odd, there is an integer k such that $m= 2k+1$, and an integer $l$ such that $n = 2l+1$. Thus it follows that $$mn= 2(2kl + k + l) + 1 $$ by concluding that $$mn= 2(2kl + k + l) + 1 \,\text{ is odd. }$$
Hence, by the equivalence of the contrapositive, we have proven: "If $mn$ is even, then either $m$ or $n$ (or both) is even."