= Solution
When $a=0$, define the <Hamiltonian function>
$$
H(x,y)=\frac{x^3}{3}+xy^2-bx.
$$
Then $\dot x=-H_y$ and $\dot y=H_x$, so $\dot H=0$. At $(\sqrt b,0)$ the Hessian of $H$ is $2\sqrt b\,I$, a strict nondegenerate minimum; at $(-\sqrt b,0)$ it is $-2\sqrt b\,I$, a strict maximum. Small regular level curves around either point are closed curves, and the <vector field> is nonzero on them. Motion around each curve therefore returns after a finite period, proving the existence of <periodic orbits> and classifying both equilibria as <center equilibria>.
The separatrix level is $H=0$, consisting of $x=0$ and $x^2/3+y^2=b$. The two arcs of this ellipse run from the lower <saddle equilibrium> to the upper one, while the vertical segment runs from the upper saddle to the lower one. They enclose the two families of <periodic orbits>, as shown in the last panel of the phase portrait. Outside the loops, regular levels are unbounded.
For $a>0$ the divergence of the original <vector field> is exactly $-a$. If a <periodic orbit> existed, uniqueness of trajectories would make it a simple closed curve, tangent everywhere to the <vector field>. The flux across its boundary would be zero, whereas the <divergence theorem> would give $-a$ times its positive enclosed area. This contradiction is the <Bendixson-Dulac criterion>. Hence $\boxed{\text{periodic orbits exist if and only if }a=0}$.
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