Let $\mathbb{V}_{1,1}$ the irreducible representation of $\mathfrak{sl}(3,\mathbb{C})$ with higest weights (1,1). I'm asked to find the decomposition in irreducibles of $S^2 (\mathbb{V_{1,1}})$ where $S^n V$ is the symmetric product of $n$ copies of $V$.
I know that $\mathbb{V}_{1,1}$ is isomorphic to the adjoint representation of $\mathfrak{sl}(3,\mathbb{C})$ i.e its weight are $(1,1),(2,-1),(-1,2),(-1,-1),(1,-2),(-2,1)$ and $(0,0)$ (with multiplicity 2).
So by taking the symmetric product i obtain a $36$-dimensional vector space.
The irreducible representation with the highest weight should be $(2,2)$, and by Weyl dimension formula the space $\mathbb{V}_{2,2}$ should be $27$-dimensional. After that the higest weight should be $(3,0)$, but this is impossible because $\mathbb{V}_{3,0}$ is $10$-dimensional and $27+10>36$.
What am i doing wrong?
Let $\alpha_1,\alpha_2$ be the roots of $\mathfrak{sl}(3)$. Now, the character of $\mathbb V_{1,1}$ is $$ f(\alpha_1,\alpha_2)=2+e^{\alpha_1}+e^{\alpha_2}+e^{\alpha_1+\alpha_2}+e^{-\alpha_1}+e^{-\alpha_2}+e^{-\alpha_1-\alpha_2}, $$ which can be viewed as a (formal) function of $\alpha_1$ and $\alpha_2$. Thus, the character of the $36$-dimensional representation $\mathrm{Sym^2}(\mathbb V_{1,1})$ is $$\begin{align*} \frac12(f(\alpha_1,\alpha_2)^2+f(2\alpha_1,2\alpha_2))=&6+3e^{\alpha_1}+3e^{-\alpha_1}+3e^{\alpha_2}+3e^{-\alpha_2}\\ &+3e^{\alpha_1+\alpha_2}+3e^{-\alpha_1-\alpha_2}+e^{\alpha_1-\alpha_2}+e^{\alpha_2-\alpha_1}\\ &+e^{2\alpha_1}+e^{-2\alpha_1}+e^{2\alpha_2}+e^{-2\alpha_2}\\ &+e^{\alpha_1+2\alpha_2}+e^{-\alpha_1-2\alpha_2}+e^{2\alpha_1+\alpha_2}+e^{-2\alpha_1-\alpha_2}\\ &+e^{2\alpha_1+2\alpha_2}+e^{-2\alpha_2-2\alpha_2}.\end{align*}$$
The highest weight is $2\alpha_1+2\alpha_2$, with representation $V(2\alpha_1+2\alpha_2)$ of dimension $27$ whose character is $$\begin{align*}&e^{2\alpha_1+2\alpha_2}+e^{-2\alpha_1-2\alpha_2}+e^{2\alpha_1}+e^{-2\alpha_1}+e^{2\alpha_2}+e^{-2\alpha_2}\\ +&e^{\alpha_1+2\alpha_2}+e^{-\alpha_1-2\alpha_2}+e^{2\alpha_1+\alpha_2}+e^{-2\alpha_1-\alpha_2}+e^{\alpha_1-\alpha_2}+e^{-\alpha_1+\alpha_2}\\ +&2e^{\alpha_1+\alpha_2}+2e^{-\alpha_1-\alpha_2}+2e^{\alpha_1}+2e^{-\alpha_1}+2e^{\alpha_2}+2e^{-\alpha_2}+3, \end{align*}$$ and the remaining character is $$\begin{align*}&3+e^{\alpha_1}+e^{-\alpha_1}+e^{\alpha_2}+e^{-\alpha_2}+e^{\alpha_1+\alpha_2}+e^{-\alpha_1-\alpha_2},\end{align*}$$ This is $9$-dimensional, and is $V(\alpha_1+\alpha_2)\oplus V(0)$. In conclusion: $$\mathrm{Sym}^2(\mathbb V_{1,1})=\mathbb V_{2,2}\oplus\mathbb V_{1,1}\oplus\mathbb V_{0,0}.$$
An easier (less computational) argument goes as follows: Observe that $\mathrm{Sym}^2(\mathbb V_{1,1})$ has weights in Weyl group orbits of $0,\alpha_1+\alpha_2,2\alpha_1+\alpha_2,2\alpha_1+2\alpha_2$. Now, we know $\mathbb{V}_{2,2}$ is $27$-dimensional so the quotient $\mathrm{Sym}^2(\mathbb V_{1,1})/\mathbb{V}_{2,2}$ is $9$-dimensional. The weights of this representation must be in the Weyl group orbit of $0$, $\alpha_1+\alpha_2$, or $2\alpha_1+\alpha_2$. However, as you observed, $V(2\alpha_1+\alpha_2)$ is $10$-dimensional, so $2\alpha_1+\alpha_2$ cannot appear as a weight. Thus, the weights are in Weyl group orbits of $0$ or $\alpha_1+\alpha_2$. We thus conclude the representation is either $V(0)^9$ or $V(0)\oplus V(\alpha_1+\alpha_2)$. However, we know the multiplicity of $V(0)$ in $V(\alpha_1+\alpha_2)^{\otimes 2}$ is $1$ (by Schur's lemma), so the quotient $\mathrm{Sym}^2(\mathbb V_{1,1})/\mathbb{V}_{2,2}$ must in fact be $V(0)\oplus V(\alpha_1+\alpha_2)$.