= Solution
Let $v_k=x^{n-k}y^k$. In the summand $V_{2n-2j}$ from part (d), a highest-weight vector is
$$
w_j=\sum_{k=0}^j(-1)^k\binom jk,v_k\otimes v_{j-k}.
$$
The raising operator $x\partial_y$ annihilates this sum because consecutive terms cancel. Interchanging the tensor factors sends $w_j$ to $(-1)^jw_j$. Since each summand in part (d) has multiplicity one, the flip acts by that scalar on the whole summand. The alternating square therefore consists exactly of the odd-$j$ summands. This <flip parity in the SU2 tensor square> gives
$$
\boxed{
\bigwedge^2V_n
\cong\bigoplus_{k=0}^{\lfloor(n-1)/2\rfloor}
V_{2n-4k-2}}.
$$
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