Note that for $V=\mathbf{R}^n$, $$S^2V = \{ v\otimes w \mid v, w\in V\text{ and }v\otimes w=w\otimes v \} =\{ A\in \mathrm{M}_2(\mathbf{R}) \mid A=A^T \}.$$ Clearly, $S^2V $ contains $O=\{ v\otimes v \mid v\in V\}$.
I want to insist that $\{ A\in S^2V \mid \det A = 0\} = O$ .
Can this statement can be found in some reference? Or this is statement wrong?
In my thought, the $n=2, 3$ cases hold.
There is an error in the introductory part of the question. Although $S^2V$ can be regarded as the space $\{A:A=A^T\}$ of symmetric matrices, the set of $v\otimes w$ with $v\otimes w=w\otimes v$ is a smaller set. It generates $S^2V$ as a vector space, but it is not equal to that vector space (unless the dimension of $V$ is very small).
Now for your actual question, notice that a symmetric matrix of the form $v\otimes v$ would have rank 1. But the $3\times 3$ matrix $\pmatrix{1&0&0\\0&1&0\\0&0&0}$ has rank 2 and its determinant is zero, so it is a counterexample to your conjecture.