Practical formula for decomposing $\textrm{Sym}^2(V_\lambda)$? Consider the irreducible $S_n$-representation $V_{\lambda}$ associated to a partition $\lambda$. In the book on Representation Theory by Fulton and Harris, there is a rather explicit formula for decomposing $V_{\lambda}\otimes V_{\lambda}$ into irreducibles (Exercise 4.5.1). Is there something similar for $\textrm{Sym}^2(V_\lambda)$? Primarily, I am interested in the multiplicity of the standard representation $V_{(n-1,1)}$ in $\textrm{Sym}^2(V_\lambda)$ but a general formula would of course be nicer.
 A: A general formula for the decomposition of $Sym^2(V_\lambda)$ seems very out of reach, but if you just care about the multiplicity of $V_{(n-1,1)}$ it's not so bad.
First let's look at the even easier case of the multiplicity of the trivial representation $\mathbf{1} = V_{(n)}$ : For every partition $\lambda$ we have that $\text{dim}(Sym^2(V_\lambda)^{S_n}) = 1$. Well since every irreducible representation of $S_n$ is self dual this means that $V_\lambda \otimes V_\lambda$ has a one dimensional space of invariants, and since each $V_\lambda$ is defined over the real numbers this means that the invariant lies in $Sym^2$ rather than $\Lambda^2$.
Okay with that in mind instead of looking at $V_{(n-1,1)}$ let's first look at the defining representation $X = V_{(n-1,1)} \oplus V_{(n)} = Ind_{S_{n-1}}^{S_n}(\mathbf{1})$. If we compute $\text{dim}(Hom(X, Sym^2(V_\lambda)))$ we can just subtract $1$ to get the answer you want. We have:
$$Hom_{S_{n}}(X, Sym^2(V_\lambda)) \cong  Hom_{S_{n-1}}(\mathbf{1}, Sym^2(Res_{S_{n-1}}^{S_n}(V_{\lambda}))) $$
This is by Frobenius reciprocity along with the fact that restriction commutes with tensor products and symmetric powers (i.e. restriction is a symmetric monoidal functor). Next let's use the branching rules for restriction from $S_n$ to $S_{n-1}$:
$$Res_{S_{n-1}}^{S_n}(V_{\lambda})= \bigoplus_{\mu = \lambda-\square} V_\mu$$
Where the sum is over all partitions $\mu$ obtained by removing a single box from the young diagram of $\lambda$. Taking the symmetric square gives:
$$Sym^2(Res_{S_{n-1}}^{S_n}(V_{\lambda}))= \bigoplus_{\mu = \lambda-\square} Sym^2(V_\mu) \oplus \bigoplus_{\mu \ne \mu'} (V_\mu \otimes V_{\mu'})$$
Hence we just need to compute the dimension of the $S_{n-1}$-invariants in this sum. This is easy though, we already saw that each $Sym^2(V_\mu)$ has a one dimensional space of invariants, and the $V_\mu \otimes V_{\mu'}$ have no invariants (the tensor product of two irreducible representations has an invariant vector iff the two representations are dual).
Putting it all together: The multiplicity of $V_{(n-1,1)}$ in $Sym^2(V_\lambda)$ is equal to the number of removable boxes in the young diagram of $\lambda$ minus $1$.
A: This is not an obvious question, but luckily it has been solved in the article "Splitting the Square of a Schur Function into its Symmetric and Antisymmetric Parts" by Carré and Leclerc in J. Alg. Comb. 1995. The rule is given in Corollary 5.5 as a counting of Yamanouchi domino tableaux defined in Section 3.

Edit:
As noted by Nate, I misread the question. I suspect it is hopeless in general, but perhaps manageable for the standard representation. I also think, taking a cue from the article by Carré and Leclerc, that figuring out which submodules come from the symmetric square versus the alternating square is harder than just decomposing the tensor product. To get started, one needs to compute the Kronecker coefficient $g_{\lambda,\lambda,(n-1,1)}$. Surprisingly (for me), I learned from slides by Greta Panova that there is formula for that due to Ballantine and Orellana. See their article in SLC 2006.
