= Solution
When $k=0$, the system is conservative and
$$
E(x,y)=\frac12y^2+\frac{x^2}{(1+x^2)^2}
$$
is constant. The fixed points are
$$
(0,0),\qquad(1,0),\qquad(-1,0).
$$
The potential $U$ has a strict minimum $U(0)=0$, strict maxima
$$
U(\pm1)=\frac14,
$$
and tends to zero as $|x|\to\infty$. The <conservative planar phase portrait> therefore has:
* a center at the origin, surrounded by closed periodic energy curves for $0<E<1/4$;
* saddles at $(\pm1,0)$;
* the level $E=1/4$ as the separatrix, including the two heteroclinic branches joining the saddles around the central periodic region;
* unbounded outer trajectories on either side of the barriers, and crossing trajectories when $E>1/4$.
The portrait is symmetric under $(x,y)\mapsto(-x,-y)$ and under time reversal $(x,y,t)\mapsto(x,-y,-t)$.
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