= Solution
Use the <Dulac theorem> on the simply connected <positive quadrant> with $\phi(x,y)=1/(xy)$. If $F=(\dot x,\dot y)$, then
$$
\phi F=
\left(\frac{1-y+ax}{y},
\frac{r(-1+x-by)}{x}\right)
$$
and hence
$$
\boxed{
\nabla\mathbin{\cdot}(\phi F)
=\frac ay-\frac{rb}{x}.}
$$
For $ab<0$, both terms have the same strict sign throughout $\Lambda$. The <Bendixson-Dulac criterion> therefore gives \b[no periodic orbit in $\Lambda$.] The same computation also excludes the case $ab=0$ unless $a=b=0$.
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