In set $\mathbb{N}$ introduce and equivalence relation such that the set $\mathbb{N}$ becomes partitioned to three classes of equivalencies, which form the corresponding quotient space.
So I’m a little confused by the question. Since by definition, every partition of $\mathbb{N}$ corresponds to an equivalence relation, is it valid to do something like $\{\emptyset\}\bigcup \mathbb{Z^+}\bigcup\{0\}$?
Not really. First and foremost, you are asked to give an equivalence relation, not a partition. True, every equivalence relation corresponds to a partition and vice versa, but they are not the same things.
Unfortunately, if you put $S_{1}=\emptyset$, $S_{2}=\mathbb{N}^{+}$, and $S_{3}=\left\{0\right\}$ and you define as usual $$x\sim y\quad\Longleftrightarrow\quad \exists 1\leq i\leq 3: x\in S_{i}\ni y$$ you get an equivalence relation with only two equivalence classes! The reason for that is that your partition is not really a partition. One of the conditions in the definition of a partition is that all sets involved are non-empty. This shows clearly why: $S_{1}$ does not lead to any equivalence class.