= Solution
A vector in a <rational representation> is an <unstable vector in a rational representation> when $0\in\overline{Gv}$, with Zariski closure. For $G=SL_n(\mathbb C)$, the <Hilbert-Mumford criterion for the affine null cone> is
$$
\boxed{v\text{ unstable}\iff\exists\lambda:\mathbb C^*\to SL_n(\mathbb C)\text{ algebraic},\quad\lim_{t\to0}\lambda(t)v=0.}
$$
Every <algebraic one-parameter subgroup> is conjugate to $\operatorname{diag}(t^{a_1},\ldots,t^{a_n})$ with integer $a_j$ summing to zero. This follows by decomposing the standard representation into characters of the multiplicative group, then imposing <determinant> one. In its weight decomposition, a limit of zero means that only strictly positive weights occur.
The reverse implication in the criterion is immediate: the subgroup lies in $G$ and supplies points of the orbit tending to zero. For the harder implication, the complex orbit is locally closed and Zariski open in its closure. A nonempty Zariski-open subset of a complex irreducible variety is dense also in the usual topology. Thus choose $g_jv\to0$ in the ordinary vector-space topology. <Singular value decomposition>, with <determinant> phases adjusted, gives
$$
g_j=k_j\operatorname{diag}(e^{b_{j1}},\ldots,e^{b_{jn}})\ell_j,\qquad k_j,\ell_j\in SU_n,\quad\sum_rb_{jr}=0.
$$
Use an $SU_n$-invariant <Hermitian inner product> on $V$. By compactness, after a subsequence $\ell_j\to\ell$. Put $w=\ell v$, and decompose it into the weights of the diagonal torus, $w=\sum_\chi w_\chi$. These <weight spaces> are orthogonal because the unitary diagonal torus preserves the inner product. Hence
$$
\|g_jv\|^2=\sum_\chi e^{2\chi(b_j)}\|(\ell_jv)_\chi\|^2\longrightarrow0.
$$
For every $\chi$ with $w_\chi\ne0$, its final norm is bounded away from zero for large $j$, forcing $\chi(b_j)\to-\infty$. There are only finitely many such weights, so choose a single real vector $b$ with sum zero and $\chi(b)<0$ for all of them. The inequalities are strict with integer coefficients. A nearby rational vector still satisfies them; multiply its negative by a common denominator to obtain integers $a_r$ of sum zero with $\chi(a)>0$ throughout this support. Then $\mu(t)=\operatorname{diag}(t^{a_r})$ takes $w$ to zero. Conjugating gives $\lambda(t)=\ell^{-1}\mu(t)\ell$ taking $v$ to zero, completing the proof. The zero vector itself satisfies the criterion trivially.
For degree-six <binary forms>, a nontrivial <algebraic one-parameter subgroup> of $SL_2$ is, after conjugation and choosing its direction, $\operatorname{diag}(t^a,t^{-a})$ with $a>0$. On $x^{6-j}y^j$ the weight is $a(6-2j)$. Strictly positive weights permit only $j=0,1,2$, so the form is divisible by $x^4$. Conversely every such form tends to zero under that subgroup. Changing basis replaces $x$ by any linear factor. The <root-multiplicity criterion for unstable binary forms> therefore gives
$$
\boxed{f\in\operatorname{Sym}^6W\text{ unstable}\iff f=0\text{ or }f\text{ has a projective root of multiplicity at least }4.}
$$
A triple root alone is insufficient: $x^3y^3$ has weight zero and is not unstable. Nor does merely having a repeated root suffice.
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