This is a bit of expansion on the answer of Mike Skirvin; in particular it gives one way of explicitly calculating the combinatorics involved. My previous answer, although correct in its result, is horribly roundabout and overly computational, a mathematical Rube Goldberg machine if you will; so after waking up this morning I realized there is a much easier approach using a recursion on Symmetric powers.

Define:

$L = U^3 + U + U^{-1} + U^{-3}$

$M = U^4 + U^2 + 2 + U^{-2} + U^{-4}$

Now define a sequence $S_i$ by:

$S_{-2} = 0$

$S_{-1} = 0$

$S_0 = 1$

$S_1 = L$

$S_k = L\cdot S_{k-1} - M\cdot S_{k-2} + L\cdot S_{k-3} - S_{k-4}$ for $k\geq 2$.

Note that the exponents of $U$ in $S_1$ are exactly the weights mentioned by Mike in his comment (1). It turns out the same is true for all the $S_k$: the exponents of $U$ in $S_k$ are exactly the set of weights of $Sym^k(Sym^3(V))$ and the coefficient of $U^\ell$ is exactly the multiplicity of the weight $\ell$ in $Sym^k(Sym^3(V))$.

From this, you can pick out the subrepresentations by looking at where coefficients change; since the weights of any $Sym^\ell(V)$ occur with multiplicity 1, the only time the coefficients change is when a new summand occurs.

For example, working out $Sym^3(Sym^3(V))$ one gets the following expression:

$U^9 + U^7 + 2U^5 + 3U^3 + 3U + 3U^{-1} + 3U^{-3} + 2U^{-5} + U^{-7} + U^{-9}$

For the module corresponding to the leading coefficient, subtract 1 from each exponent giving a copy of $Sym^9(V)$ and leaving:

$U^5 + 2U^3 + 2U + 2U^{-1} + 2U^{-3} + U^{-5}$

Repeat this process to pull out a copy of $Sym^5(V)$ and finally a copy of $Sym^3(V)$; there are no more terms left, so this is the complete decomposition of $Sym^3(Sym^3(V))$. In general, the expression for $Sym^k(Sym^3(V))$ in $U$ so obtained is of the form:

$a_0U^{3k} + a_2U^{3k-2} + a_4U^{3k-4} + ... + a_4U^{-3k+4} + a_2U^{-3k+2} + a_0U^{-3k}$

Then $a_0 = 1$ by Mike's comment (2) and the multiplicity of $Sym^\ell(V)$ for $\ell\geq 0$ in the decomposition is just $(a_\ell - a_{\ell+2})$ and the multiplicity of $Sym^{3k}(V)$ is 1 since $a_{-2} = 0$.

As for the recursion, it ultimately expresses symmetric powers in terms of lower symmetric powers and exterior powers; this can be proven using multiplication of Young diagrams and inclusion-exclusion although I don't have a good reference at hand.