Yes, this proof is strange.

It is not clear immediately why q is odd.This is clearly non sequitur. Not every odd number is a multiple of 3.

One proof that is irrational parallels a similar proof for . Suppose 3q^2 = p^2 where GCD(p, q) = 1. Then 3 | p^2, so by Euclid's lemma, 3 | p, i.e., p = 3k for some integer k. Then 3q^2 = 9k^2, so 3 | q^2 and 3 | q, which is a contradiction with GCD(p, q) = 1.