The following relation is defined on $\mathbb{N}^{+}$
True or false? This relation is an equivalence relation: $xRy \Leftrightarrow x \cdot y$ is even
First, relation is equivalence relation if it have property: reflexive, symmetric, transitive.
$xRy \Leftrightarrow x \cdot y$ is even $\text{ }\text{ }\text{ }\text{ }$ This says that when product is even, you have relation that satisfy all these property i.e. is equivalence relation, right?
Is allow I negate the relation and find counterexample?
Negation say: If product of $x,y$ is uneven, you don't get equivalence relation.
Let me make example, $x=y=1:$ $1 R 1 \Leftrightarrow 1$ is uneven
But $(1,1)$ is equivalence relation. Statement is false.
Is all good work or I do wrong?
$R$ is a relation. A relation can either be True or False for a pair. So when we look at $1R1$ this state that $1*1$ is even, which is clearly false, just like you have stated.
As you say, an equivalence relation has to be reflexive i.e. for each positive integer x, $xRx$ should hold. But you have shown that $1R1$ does not hold, and thus $R$ is not reflexive and thus $R$ is not an equivalence relation.
Your proof is correct and good up to the English. It would also be nice to see that you recognize that you are contradicting the reflexivity.