I guess this question only requires standard knowledge, but I'm a bit rusty with highest weight theory. I'm trying to catch up, but maybe I don't need the theory in full generality.
Background
Let $V$ be a Euclidean space of dimension $n$, and consider the representations of the group $G = O(V) \cong O(n)$. If I'm not wrong we can decompose $$ \operatorname{Sym}^2 V = W \oplus Z, $$ where $Z$ is the trivial one-dimensional representation and $W$ is an irreducible representation of $G$.
More background
This is how I can see the above decomposition. I don't know if it is of any use for the question itself, so feel free to skip it.
It is enough to decompose $\operatorname{Sym}^2 V^{*}$, that is, degree $2$ homogeneous polynomials on $V$. Degree $n$ homogeneous polynomials are a representation of $G$ via $$ A.f(x) = f(Ax), $$ where $A$ is an orthogonal matrix and $f \in \operatorname{Sym}^n V^{*}$.
Now for any $n$ we have a $G$-morphism $f : \operatorname{Sym}^n V^{*} \to \operatorname{Sym}^{n + 2} V^{*}$ which is given by multiplication by $x^2 = x_1^2 + \cdots + x_n^2$. This exhibits $\operatorname{Sym}^n V^{*}$ as a subrepresentation of $\operatorname{Sym}^{n + 2} V^{*}$.
The complement is easily found. The invariant scalar product on $\operatorname{Sym}^n V^{*}$ is given by $(f, g) = f(D)g(x)$, where $D$ is the derivation operator. From this one finds that the adjoint of $f$ is the laplacian $\Delta$.
So one can decompose $$ \operatorname{Sym}^{n + 2} V^{*} = \operatorname{Sym}^{n} V^{*} \oplus \mathcal{H}_n, $$ where $\mathcal{H}_n$ is the space of harmonic homogeneous polynomials of degree $n$. If I recall well, $\mathcal{H}_n$ is irreducible.
The decomposition which interests me is then $$ \operatorname{Sym}^{2} V^{*} = \mathcal{H}_2 \oplus \mathcal{H}_0. $$
Problem
I'd like to understand how to decompose the tensor products of $W$; in particular
What is the decomposition of $W \otimes W$ and $W \otimes W \otimes W$ into irreducible representations of $G$?