= Solution
Let $A(t)=D_{(x,v)}(X,V)$ along one <characteristic curve>. Differentiating the characteristic equation gives $\dot A=D_zb(t,X,V)A$, $A(0)=I_6$, with $b=(V,F)$. The <Liouville formula for a fundamental matrix> then gives the <phase-space flow Jacobian>
$$
\boxed{J(t,x,v)=\exp\left(\int_0^t\nabla_v\cdot F(s,X(s),V(s))\,ds\right)}.
$$
Indeed $\operatorname{tr}D_zb=\operatorname{div}_{x,v}(v,F)=\nabla_v\cdot F$, since $v$ is independent of $x$. Differentiation proves $\dot J=(\nabla_v\cdot F)J$. If that <divergence> vanishes, $J(0)=1$ gives $\boxed{J\equiv1}$ and the flow preserves <Lebesgue measure> on <phase space>.
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