= Solution
The <Poisson bracket> is $\{f,g\}=\mathbf r\cdot(\nabla f\times\nabla g)$. Since $\nabla\rho^2=2\mathbf r$, the <scalar triple product> vanishes, and
$$
\boxed{\{f,\rho^2\}=0.}
$$
Equivalently, check the identity on $x,y,z$ and extend to every <polynomial> using the <Leibniz rule>. Thus $\rho^2$ is a <Casimir function of a Poisson manifold> of this <rotational Lie-Poisson structure on R3>.
The <Hamilton's equations> are
$$
\boxed{\dot x=(B-C)yz,\qquad\dot y=(C-A)xz,\qquad\dot z=(A-B)xy.}
$$
<Linearization> about $(1,0,0)$ gives $\dot\alpha=0$, $\dot\beta=(C-A)\gamma$, $\dot\gamma=(A-B)\beta$. The two nonzero exponents satisfy $\lambda^2=(C-A)(A-B)$. Consequently
$$
\boxed{\text{exponential instability occurs iff }\min(B,C)<A<\max(B,C).}
$$
When $A$ is outside this interval the transverse motion is oscillatory. At equality the <linearization> is nonhyperbolic; if just one transverse coupling vanishes, the other permits secular linear growth, rather than an exponential instability. This is the familiar intermediate-axis instability described by the <Euler equations for a torque-free rigid body>.
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