= Solution
The <Bendixson-Dulac theorem> says that a continuously differentiable planar <vector field> has no nonconstant <periodic orbit> in a <simply connected> region if a continuously differentiable <Dulac function> makes its rescaled <divergence> have one sign and not vanish identically on any open subset of that region. Isolated zeros of the <divergence> are allowed in this form.
Take the <Dulac function> to be one. For the exact <vector field> $(P,Q)$,
$$
\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}
=3x^2+7y^2+12x^2y^2.
$$
This is positive except at the origin. The plane is <simply connected>, so \b[there are no nonconstant periodic solutions]. Equivalently, a hypothetical closed trajectory would have zero outward flux of its tangent <vector field>, while <Green's theorem> would make that flux the strictly positive area <integral> of the <divergence>.
The origin is the only <equilibrium point>: $Q=0$ forces $y=0$, and then $P=x^3$ forces $x=0$. Its <Jacobian matrix> is zero, so the linearized equations are $\dot x=\dot y=0$. They give no stability classification.
Keeping cubic terms gives the homogeneous approximation
$$
\dot x=x(x^2+y^2),\qquad\dot y=2y^3.
$$
In a sector where $xy\ne0$, use $y$ as the independent variable and put $w=x^{-2}$. Dividing the two equations gives
$$
\frac{dx}{dy}=\frac{x(x^2+y^2)}{2y^3},\qquad
\frac{dw}{dy}+\frac wy=-\frac1{y^3}.
$$
The <integrating factor> $y$ gives $(yw)'=-y^{-2}$, hence $yw=1/y+C_0$. Rearranging and renaming the arbitrary constant yields the implicit trajectories
$$
\boxed{\frac1y-\frac y{x^2}=C,\qquad\text{equivalently}\qquad x^2-y^2=Cx^2y.}
$$
This is the <first integral> of the cubic approximation, not an exact first integral of the higher-degree original system. The invariant axes, excluded by the divisions, must be added: on $y=0$, $\dot x=x^3$; on $x=0$, $\dot y=2y^3$. The diagonal trajectories $y=\pm x$ correspond to $C=0$. For fixed finite $C$, non-axis trajectories approaching the origin satisfy $y^2/x^2=1-Cy\to1$, so their limiting tangent lines are the diagonals. This is the <cubic planar flow with diagonal approach directions>.
Every nonzero trajectory moves outward locally. For $V=x^2+y^2$, the exact system gives
$$
\dot V=2x^4+2x^2y^2+4y^4+2x^4y^2+6x^2y^4\geq V^2>0.
$$
This implies departure from any fixed small circle for every nonzero starting point inside it: while a trajectory stayed inside forever, $\frac{d}{dt}(1/V)\leq-1$ would eventually force $1/V$ negative. Thus the origin is \b[unstable and repelling] for both the cubic approximation and the full system. The argument establishes nonlinear instability despite the zero linearization.
\Image[/past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2014-cubic-origin-trajectories.png]
{title=Implicit trajectories of the cubic approximation: invariant axes and diagonal limiting directions, with all forward arrows outward}
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