= Solution
Use the <Clebsch-Gordan decomposition for SU2>, which states that $V_m\otimes V_n\cong\bigoplus_{j=0}^{\min(m,n)}V_{m+n-2j}$. Here this gives
$$
\boxed{V_3\otimes V_3\cong V_6\oplus V_4\oplus V_2\oplus V_0.}
$$
To determine symmetry, realize the tensor product as polynomials in two pairs of variables. A highest-weight vector in the $j$th summand can be chosen proportional to $(x_1y_2-y_1x_2)^jx_1^{3-j}x_2^{3-j}$. Exchanging the two factors multiplies it by $(-1)^j$, and hence multiplies the entire irreducible summand by that sign. Even $j$ belong to the <symmetric square>, odd $j$ to the <exterior square>. Thus
$$
\boxed{S^2V_3\cong V_6\oplus V_2,\qquad\bigwedge^2V_3\cong V_4\oplus V_0.}
$$
Their dimensions are respectively $7+3=10$ and $5+1=6$, as required for a four-dimensional space.
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