I have the following system of differential equations: $$\begin{cases}f(x,y)=x(1-x-y)\\ g(x,y)=y(1-x-y)\end{cases}$$
I want to see that there are no periodic orbits in the first quadrant. First, note that the lines $x=0$, $y=0$ are two nullclines and the movement is only vertical and horizontal respectively.
Now, taking the Dulac function $\phi(x,y)=\frac{1}{xy}$ I have that $$\frac{\partial f(x,y)\phi(x,y)}{x}+\frac{\partial f(x,y)\phi(x,y)}{y}=\frac{-1}{y}+\frac{-1}{x}<0 \text{ for all }x,y>0$$
Using the Bendixson-Dulac theorem, the system can't have periodic orbits in the first quadrant. The theorem doesn't deny that it could exist a periodic solution that crosses the lines $x=0$,$y=0$, however since the lines $x=0$, $y=0$ are two nullclines it can't happen and therefore we can't have periodic solutions in the first quadrant ($x,y\geq 0$).
Is this correct?
Hint.
Note that $\frac{dy}{dx} = \frac yx$ is a movement integral so the orbits are of type $y = \lambda x$ then no closed orbits are possible.