 Solution

Use the Dulac theorem on the simply connected positive quadrant with $\varphi (x,y)=1/(xy)$. If $F=(\dot{x},\dot{y})$, thenand henceFor $ab<0$, both terms have the same strict sign throughout $\Lambda$. The Bendixson-Dulac criterion therefore gives no periodic orbit in $\Lambda$. The same computation also excludes the case $ab=0$ unless $a=b=0$.