= Solution
The <symmetric group> $S_3$ has three <conjugacy classes>, determined by cycle type: the identity, three <transpositions>, and two three-cycles. Hence it has three complex <irreducible representations>. Two are the <trivial representation> $\mathbf1$ and the <sign representation> $\varepsilon$. Their <dimensions> are one, so the <sum of squares of irreducible representation dimensions> gives $1+1+d^2=6$, whence $d=2$ for the remaining <representation> $D$.
Write $\chi_D=(2,a,b)$ on these classes. Orthogonality to $\mathbf1$ and $\varepsilon$ gives $2+3a+2b=0$ and $2-3a+2b=0$. Thus $a=0,b=-1$, and the <character table> is
$$
\boxed{\begin{array}{c|rrr}
&e&(12)&(123)\\\hline
\text{class size}&1&3&2\\\hline
\mathbf1&1&1&1\\
\varepsilon&1&-1&1\\
D&2&0&-1
\end{array}}
$$
The <character> of a <tensor product representation> is the product of the factor <characters>. Thus $\varepsilon^2=\mathbf1$, $\varepsilon\otimes D=D$, and $\chi_{D\otimes D}=(4,0,1)=\chi_{\mathbf1}+\chi_\varepsilon+\chi_D$. These determine all the <Clebsch-Gordan decomposition of group representations>:
$$
\boxed{\begin{array}{c|ccc}
\otimes&\mathbf1&\varepsilon&D\\\hline
\mathbf1&\mathbf1&\varepsilon&D\\
\varepsilon&\varepsilon&\mathbf1&D\\
D&D&D&\mathbf1\oplus\varepsilon\oplus D
\end{array}}
$$
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