So set of size $2n$ can be divided in two equal sets of size $n$, A and B. There are four ways to choose 3 elements from those 2. First two are to choose all 3 elements from either one, $2{n\choose3}$. Other two are to choose 2 from one and 1 from another, $n^2(n-1)$. That is if we choose 2 from A and 1 from B, but it can also be done the other way around, therefore number of ways is $2n^2(n-1)$.
Finaly $$ {2n \choose 3} = 2{n \choose 3} + 2n^2(n-1) $$
but in the problem it is just $n^2(n-1)$.
Am I doing it wrong?
The proof idea is correct, well done.
When you are saying "choose two elements from $A$ and one element from $B$", you write the answer as $n^2(n-1)$, which is incorrect, as choosing two elements from $A$ is done in $\binom n2$ ways, while you seem to not have divided by $2$ in the previous expression. A similar thing has happened when you took two elements from $B$ and $1$ from $A$. So, the actual answer got doubled because you forgot the half coming from the binomial.
Set this right, and you get your result.