= Solution
Part d leaves only the region above $a=-1$ and to the right of $b=-1$ lying below the hyperbola $ab=1$. Part c removes the two opposite-sign quadrants, and its displayed divergence also removes either coordinate axis except at the origin. Hence periodic orbits might still occur only in
$$
\boxed{
\{a>0, b>0, ab<1\}
\;\cup\;
\{-1<a<0, -1<b<0\}
\;\cup\;\{(0,0)\}.}
$$
In the $a$-$b$ plane these are the portion of the first quadrant below $ab=1$, the open square $(-1,0)\times(-1,0)$, and the origin. The origin is not merely possible: part a gives its family of periodic orbits.
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