Note the following.

For all sets X and Y, x belongs to X ∆ Y iff x belongs to precisely one of X and Y; correspondingly, x does not belong to X ∆ Y iff x belongs to either both X and Y or none of them. (*)

There are 2³ = 8 cases for whether x belongs to each of A, B and C. You can decide to be patient and go through all of them checking for each case whether x belongs to either side of the equality using (*).

Another way is to prove first that x belongs to (A ∆ B) ∆ C iff x belongs to an odd number of sets (either one or three). For any set X, define in(x, X) = 1 if x ∈ X and in(x, X) = 0 if x ∉ X. Then we want to prove the following.

For all sets A, B and C, x ∈ (A ∆ B) ∆ C iff in(x, A) + in(x, B) + in(x, C) is odd. (**)

Note that (*) says the same thing, but about two instead of three sets. To show (**), note that x ∈ (A ∆ B) ∆ C iff either (1) x ∈ A ∆ B and x ∉ C or (2) x ∉ A ∆ B and x ∈ C. In (1), in(x, A) + in(x, B) is odd and in(x, C) = 0, so in(x, A) + in(x, B) + in(x, C) is odd. In (2), in(x, A) + in(x, B) is even and in(x, C) = 1, so again in(x, A) + in(x, B) + in(x, C) is odd. (Note that this proof of (**) can be used in a proof by induction of the generalization of (**) for n sets.)

Now, given specific sets A', B', C', we can instantiate A = B', B = C' and C = A' in (**) to get

x ∈ (B' ∆ C') ∆ A' iff in(x, B') + in(x, C') + in(x, A') is odd. (***)

But (B' ∆ C') ∆ A' = A' ∆ (B' ∆ C'), so (***) gives

x ∈ A' ∆ (B' ∆ C') iff in(x, A') + in(x, B') + in(x, C') is odd, (****)

which together with (**) proves associativity. Alternatively, (****) can be shown similarly to (**).

See also this thread.