= Solution
Take $\theta=0$ initially. The <velocity> components are
$$
\dot r(0)=-4\cos(\pi/3)=-2,
\qquad r\dot\theta(0)=4\sin(\pi/3)=2\sqrt3,
$$
so $l=2\sqrt3$. Since $f(1/u)=3u^2+9u^3$, the orbit equation becomes
$$
u''+\frac14u=\frac14.
$$
Thus
$$
u=1+A\cos(\theta/2)+B\sin(\theta/2).
$$
The data $u(0)=1$ and $\dot r=-lu'$ give $A=0$, $B=2/\sqrt3$, hence
$$
\boxed{u(\theta)=1+\frac2{\sqrt3}\sin(\theta/2)}.
$$
At $\theta=2\pi$, $u=1$ again, so the particle returns to its initial position after one revolution; now $u'=-1/\sqrt3$, so it is moving outward. Subsequently $u$ first reaches zero at $\theta=8\pi/3$, where $r=1/u\to\infty$. It therefore flies off to infinity.
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