= Solution
The variables commute, so $x_ix_j=x_jx_i$. Thus $V$ has <dimension> six, with <basis> $x_1^2,x_2^2,x_3^2,x_1x_2,x_1x_3,x_2x_3$, even though nine coefficients occur in the redundant expression in the source. Its <representation> is the <symmetric square> of the three-dimensional <permutation representation>.
The identity fixes all six <monomials>. A three-cycle fixes none. For a <transposition>, say $(12)$, exactly $x_3^2$ and $x_1x_2$ are fixed; the remaining <monomials> occur in exchanged pairs. Therefore $\chi_{\widetilde R}=(6,0,2)$ in the <conjugacy class> order of <2/b/solution>. The <character inner products> give multiplicities
$$
n_{\mathbf1}=\frac{6+3\cdot2}{6}=2,\qquad
n_\varepsilon=\frac{6-3\cdot2}{6}=0,\qquad
n_D=\frac{6\cdot2}{6}=2.
$$
Consequently
$$
\boxed{\widetilde R\cong2\mathbf1\oplus2D.}
$$
There is also a concrete decomposition: the span of the three squares is one copy of the <permutation representation>. The span of the three mixed <monomials> is another, since $x_2x_3,x_1x_3,x_1x_2$ are permuted exactly like their omitted indices $1,2,3$. Within each, the coefficient-sum-zero <vector subspace> gives a copy of $D$, and the sum of its three <basis> <monomials> gives a <trivial representation>.
A fixed <quadratic polynomial> must have equal coefficients on each of the two <orbits> of <monomials>. Hence its fixed <vector subspace> is
$$
\boxed{V^{S_3}=\operatorname{span}_{\mathbb C}\{x_1^2+x_2^2+x_3^2,\quad x_1x_2+x_1x_3+x_2x_3\}.}
$$
The two displayed <polynomial invariants> are linearly independent and account for both singlets. Equivalently, one may use $x_1^2+x_2^2+x_3^2$ and $(x_1+x_2+x_3)^2$ as a <basis>. No constant term belongs to this homogeneous degree-two <vector space>.
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