= Solution
Let $k>2$ and choose the positive <Dulac function>
$$
\psi(x,y)=\exp\!\left(-\frac{2(x+y)}{k-2}\right).
$$
More generally, for $\psi=e^{a(x+y)}$, direct differentiation gives
$$
\nabla\cdot(\psi(X,Y))=\psi\bigl[3+(-2+(2-k)a)x-ax^2\bigr].
$$
Taking $a=-2/(k-2)$ cancels the linear term and makes the quadratic term nonnegative:
$$
\boxed{\nabla\cdot(\psi(X,Y))=\psi\left(3+\frac{2x^2}{k-2}\right)>0.}
$$
The open positive quadrant is <simply connected>, so the <Bendixson-Dulac theorem> excludes every nonconstant <periodic orbit> entirely inside it. Such an orbit cannot touch $x=0,y>0$ while staying in the closed quadrant, since its differentiable coordinate $x$ would have a local minimum there but $\dot x=-y\ne0$. The same argument applies on $y=0,x>0$. It cannot pass through the origin either, by local uniqueness of this smooth <autonomous system>. Therefore \b[there is no nonconstant closed trajectory in the entire nonnegative quadrant]. The calculation illustrates an <exponential Dulac multiplier for a quadratic planar system>.
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