= Solution
Let $\alpha_i^\vee=2\alpha_i/(\alpha_i,\alpha_i)$ be the <coroots>. With a chosen set of <simple roots>, the <root lattice> and <weight lattice> are
$$
\boxed{Q=\bigoplus_{i=1}^r\mathbb Z\alpha_i,\qquad P=\{\lambda:\langle\lambda,\alpha_i^\vee\rangle\in\mathbb Z\text{ for every }i\}.}
$$
The <fundamental weights> $\omega_j$ are defined by $\langle\omega_j,\alpha_i^\vee\rangle=\delta_{ij}$, and form an integral basis of $P$. We use the <Cartan matrix> convention $A_{ij}=\langle\alpha_j,\alpha_i^\vee\rangle$.
For the <A2 root system>, inversion of its <Cartan matrix> yields the <A2 fundamental weights and weight lattice>:
$$
\boxed{\omega_1=\frac{2\alpha_1+\alpha_2}{3},\qquad\omega_2=\frac{\alpha_1+2\alpha_2}{3},\qquad P=\mathbb Z\omega_1\oplus\mathbb Z\omega_2.}
$$
Equivalently $\alpha_1=2\omega_1-\omega_2$ and $\alpha_2=-\omega_1+2\omega_2$. The <root lattice> has index three in the <weight lattice>, since the change-of-basis matrix has determinant three. For a planar realization take
$$
\alpha_1=(1,0),\quad\alpha_2=(-\tfrac12,\tfrac{\sqrt3}{2}),\quad\omega_1=(\tfrac12,\tfrac{\sqrt3}{6}),\quad\omega_2=(0,\tfrac{\sqrt3}{3}).
$$
The <weight lattice> is triangular. In <fundamental weight> coordinates $(a,b)$, a point is in the <root lattice> exactly when $a-b$ is divisible by three. The following sketch marks both bases and the sublattice:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-302-a2-weight-lattice.png]
{title=A2 weight lattice, root sublattice, fundamental weights and simple roots}
The integers called Dynkin indices here are the <Dynkin labels> of the <highest weight>:
$$
\boxed{\Lambda_i=\langle\Lambda,\alpha_i^\vee\rangle,\qquad\Lambda=\sum_i\Lambda_i\omega_i.}
$$
For finite-dimensional <Irreducible Lie algebra representations>, they are nonnegative integers. All weight coordinates below are these <Dynkin label> coordinates, not simple-root coordinates.
A <weight-string enumeration algorithm> gives the set of weights without their <weight multiplicities>. Begin with $\Lambda$ and process known weights by increasing height below $\Lambda$. For each simple root and known weight $\mu$, find the largest $p\geq0$ with $\mu+p\alpha_i$ a weight; all such higher weights have already been processed. The <weight string> theorem says that the string has endpoints $\mu+p\alpha_i$, $\mu-q\alpha_i$, with
$$
q-p=\langle\mu,\alpha_i^\vee\rangle,
$$
and includes every intermediate step. Append $\mu-\alpha_i,\ldots,\mu-q\alpha_i$ and repeat until no new weights appear. This terminates in finite dimension and supplies all weights; every nonhighest weight can be reached by simple-root lowering. A string can contain contributions from several sl2 summands, so the resulting set does not by itself determine <weight multiplicities>.
For the explicit calculation we use the following general facts: finite-dimensional representations of a complex <semisimple Lie algebra> are completely reducible by the <Weyl complete reducibility theorem>; the <symmetric powers of the defining sln representation> are irreducible of highest weight $m\omega_1$; and weights in a <tensor product of Lie algebra representations> add with multiplicities multiplied. We also use the <highest-weight representation> classification: each finite-dimensional irreducible has a unique dominant integral <highest weight>, and a nonzero vector killed by all simple-root raising operators supplies an irreducible summand with that highest weight in a completely reducible module. For $A_2$, the <Weyl dimension formula> specializes to
$$
\dim R(p,q)=\frac12(p+1)(q+1)(p+q+2).
$$
For $A_2\cong\mathfrak{sl}_3(\mathbb C)$, the defining <fundamental representation> has weights $(1,0),(-1,1),(0,-1)$. Thus the <A2 representation of highest weight (2,0)> is $\operatorname{Sym}^2\mathbb C^3$, with the six distinct weights
$$
\boxed{W_6=\{(2,0),(0,1),(1,-1),(-2,2),(-1,0),(0,-2)\}.}
$$
Each has multiplicity one: these are the weights of the six quadratic monomials.
For the <tensor square of the A2 representation of highest weight (2,0)>, add every ordered pair of elements of $W_6$. To list the complete answer compactly, define four disjoint weight sets:
$$
C=\{(4,0),(-4,4),(0,-4)\},
$$
$$
E=\{(2,1),(3,-1),(-2,3),(-3,2),(1,-3),(-1,-2)\},
$$
$$
M=\{(0,2),(2,-2),(-2,0)\},\qquad I=\{(1,0),(-1,1),(0,-1)\}.
$$
The tensor product has multiplicity one at each point of $C$, two at each point of $E$, three at each point of $M$, and four at each point of $I$. These fifteen distinct weights account for $3+12+9+12=36$ states.
We next identify the irreducible summands rather than just their dimensions. Split the <tensor square> of the six-dimensional space into its <symmetric power> and <exterior power>, of dimensions $21$ and $15$. The square of a highest-weight vector in the symmetric part has <highest weight> $(4,0)$, giving $R(4,0)$ of dimension $15$.
Let $v_a,v_b$ be vectors of weights $a=(2,0)$ and $b=(0,1)=a-\alpha_1$. In the exterior part $v_a\wedge v_b$ has weight $(2,1)$ and is killed by both simple-root raising operators: raising $v_b$ along $\alpha_1$ gives a multiple of $v_a$, whose wedge with itself is zero, and the other raising actions vanish. Thus it is a highest-weight vector. The <Weyl dimension formula> gives $\dim R(2,1)=15$, so the entire exterior part is this irreducible summand.
The remaining symmetric part has dimension six. To identify it, the fifteen weights of $R(4,0)=\operatorname{Sym}^4\mathbb C^3$ are exactly
$$
(a-b,b-c),\qquad a,b,c\geq0,\quad a+b+c=4,
$$
each once. Subtract them from the unordered-pair weights of $\operatorname{Sym}^2R(2,0)$. The residual weights are $M\cup I$, each once, with highest weight $(0,2)$. They are the negatives of $W_6$, so this six-dimensional summand is $R(0,2)$. Consequently
$$
\boxed{R(2,0)\otimes R(2,0)=R(4,0)\oplus R(2,1)\oplus R(0,2).}
$$
The corresponding dimensions are $15+15+6=36$; complete reducibility and the exhibited highest weights ensure that no summands are missing.
The full <weight multiplicities in the A2 tensor square of highest weight (2,0)> are summarized below. An entry is the multiplicity of each individual weight in that row's set:
$$
\begin{array}{c|rrrr}
\text{weight set}&R(2,0)\otimes R(2,0)&R(4,0)&R(2,1)&R(0,2)\\\hline
C&1&1&0&0\\
E&2&1&1&0\\
M&3&1&1&1\\
I&4&1&2&1
\end{array}
$$
Thus $R(4,0)$ has all fifteen weights once, $R(0,2)$ has the six weights $M\cup I$ once, and $R(2,1)$ has its nine boundary weights $E\cup M$ once and its three interior weights $I$ twice. \b[Only the three interior weights of $R(2,1)$ are degenerate among the irreducible components.]
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