The Relation: $\left\{R = ((m, n) |\ mn \geq 0\ \right\} on \ \mathbb{Z}$ apparently has an equivalence class.
I can't really see it, I can see that reflexive does not fail. From the looks for it, symmetry does not fail either... But I can see transitivity fails at least from my view.
If you have: $(-1, 0)$ and $(0, 1)$, thus $(-1, 1)$ which shows that it fails since $-1 \geq 0$ is not true, which shows transitivity fails, does it not?
Denote the equivalence of $a$ and $b$ as $a \sim b$. To show that $R$ is not an equivalence relation, you need to show that one of the reflexive, symmetric, or transitive properties does not hold. You did this correctly by identifying the similarity of $(-1,0)$ and $(0,1)$, and showing that this breaks transitivity with $(-1,1)$.
The question would be more interesting if we had $mn >0$ instead of $mn \geq 0$, i.e. $R = \{(m,n) \mid mn > 0 \}$ for $m,n \in \mathbb{Z}$.
In that case, reflexivity and symmetry are obviously true. (Prove them yourself if necessary.) So suppose $a\sim b$, so $ab > 0$, and suppose $b \sim c$, so $bc >0$. We WTS $a \sim c$, i.e. $ac > 0$.
We note that if $a \sim b$, then we have one of two cases: either $a>0$ and $b>0$ or $a<0$ and $b<0$. Suppose the latter. if $a<0$ and $b<0$, but also $bc>0$, then this implies that $c<0$ as well, as if it were the case that $c \geq0$, then $bc \leq0$, and we couldn't have that. And given $a<0$ and $c<0$, then $ac >0$. The other case is if $a>0$ and $b>0$. This is similar to the previous case: if $b>0$ and $bc >0$, then $c>0$ as well, as we couldn't have $c \leq 0$ without $bc \leq 0$. So given $a >0$ and $c>0$, we have $ac >0$. So if we had $mn >0$ instead of $mn \geq 0$, R would be an equivalence relation.