I simply don't understand the question. Doesn't $\mathbb{Z}/42\mathbb{Z}$ consist of $42$ subsets? (Namely $\{0,42,84,\cdots\}, \{1,43,85,\cdots\}, \cdots, \{41, 83, 125, \cdots\}$) And then how would you square an entire subset?
New to this website so if I did anything wrong please tell me.
@TeresaLisbon is right. This is a question about modulo $42$ residue classes. It asks how many integers satisfy $0\le x\le 41,\,42|x^2-x$.
You are correct, each element of the ring $\mathbb{Z}/42\mathbb{Z}$ is strictly speaking a "coset" - which is a set as you have written. One conviniently thinks about it however as "remainders under division by $42$". So we can choose a "representative" for each element from the set $\{0, ..., 41 \}$
There are two operations to be aware of in $\mathbb{Z}/42\mathbb{Z}$, addition an multiplication - and they occur as you would expect with the above interperatation. Compute the sum (resp. product) in $\mathbb{Z}$ and compute residue. Therefore e.g., $7^2 = 49 \equiv 7 \pmod{42}$.
To your main question, you could simply iterate all of the elements in $\mathbb{Z}/42\mathbb{Z}$. I'll give a different method.
By the Chinese Remainder Theorem $$\mathbb{Z}/42\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z}$$
Then some $a \in \mathbb{Z}/42\mathbb{Z}$ satisfies $a^2 \equiv a \pmod{42}$ if and only if $a^2 \equiv a \pmod{7}$ and $a^2 \equiv a \pmod{2}$ and $a^2 \equiv a \pmod{3}$. Thus it occurs if and only if $a \equiv 0, 1 \pmod{7}$ and $a \equiv 0, 1 \pmod 3$ and $a \equiv 0, 1 \pmod 2$.
Thus the elements of $\mathbb{Z}/42\mathbb{Z}$ satisfying the relation are $0, 7, 21, 28, 36, 1, 15, 22$
