 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 inIn 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.