You are correct that the relation defined is an equivalence relation on A is an equivalence relation, essentially $$\forall x \in A, xRy \iff 3\mid (x - y) \iff x\equiv y \pmod 3$$
The relation $R$, i.e., defines congruence modulo $3$. So your task boils down to finding the congruence classes, $\pmod 3$.
Do you know how to find the equivalence classes of your set, $\pmod 3$?
- Class one: Which elements have are divisible by $3$? $\quad A_0 = \{-3, 0, 3\}$
- Class two: Which elements leave a remainder of $1$ when divided by $3$? $\quad A_1 =\{-2, 1, 4\}$
- Class three: Which elements leave a remainder of $2$ when divided by $3$? $\quad A_2 = \{-4, -1, 2, 5\}$
You're done: three equivalence classes.
$$A = A_0 \cup A_1 \cup A_2 = \{-4,-3,-2,-1,0,1,2,3,4,5\},$$ $$ \quad A_i \cap A_j = \varnothing, \;\text{when}\;\;i\neq j, \text{ for}\;\; i, j \in \{0, 1, 2\}$$
Two elements $(a,b)$ and $(c,d)$ in $S \times S$ are related if $a+b=c+d$. For example, if we take the element $(1,5) \in S \times S$, then $(1,5) \sim (1,5)$ because $1+5=1+5=6$.
So for finding the equivalence class of $(1,5)$ we ask ourselves: what are all other elements $(c,d) \in S \times S$ such that $(1,5) \sim (c,d)$?
For that, we want $c+d=6$. So look for all the pairs that satisfy this condition.
$$[(1,5)]=\{(1,5), (5,1), (3,3), (2,4), (4,2)\}.$$
Likewise
$$[(1,1)]=\{(1,1)\} \qquad \text{and} \qquad [(5,6)]=\{(5,6), (6,5)\}.$$
Hopefully you can proceed from here to get the remaining equivalence classes.
Note: If you just want the number of equivalence classes (without describing them), then note that each equivalence class can be associated with the sum of the pairs in that, e.g. the class $[(1,5)]$ can be associated to the sum $6$ and class $[(1,1)]$ can be associated with the sum $2$ and so on. So the number of distinct classes is the number of distinct sums.
Best Answer
Your intuition is correct. To describe the equivalence classes explicitly, it often helps to find the equivalence classes of numbers that are easy to work with.
Let's find, for example, the equivalence classes $[0]$ and $[1]$.
$0+a$ is even for which integers $a$? All these will form $[0]$.
$1+b$ is even for which integers $b$? All these will form $[1]$.
Are there any other equivalence classes? (You can be sure you've found them all when your equivalence classes form a partition of $\Bbb Z$.)