= Solution
Completing squares reveals the geometry of the <sheared quartic double-well gradient flow>:
$$
V=(x^2-1)^2+(x+y)^2-1.
$$
The <critical points> satisfy $2(x+y)=0$ and $4x^3-2x+2y=0$, so they are $(0,0)$, $(1,-1)$ and $(-1,1)$. The <Hessian matrix> is
$$
\operatorname{Hess}V=\begin{pmatrix}12x^2-2&2\\2&2\end{pmatrix}.
$$
At the origin its determinant is $-8$, so $(0,0)$ is a <saddle point>, with $V=0$. At either other <critical point>, the determinant is $16$ and the upper-left entry is $10$, so it is a <positive-definite matrix>. They are strict <local minima>; the completed squares show that both are global minima, with $V=-1$.
A contour at level $c$ obeys
$$
(x+y)^2=c+1-(x^2-1)^2,\qquad
y=-x\pm\sqrt{c+1-(x^2-1)^2}.
$$
There are no contours for $c<-1$, two isolated minimum points for $c=-1$, and two separate closed ovals for $-1<c<0$. At $c=0$ the two lobes meet at the <saddle point>; near the origin their tangents are $y=(-1\pm\sqrt2)x$. For $c>0$ one closed contour surrounds both minima.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/ia/paper-2-double-well-contours.png]
{title=Sheared quartic potential contours and the trajectory to the positive-x minimum}
Along the <gradient flow>, the <gradient-flow dissipation identity> is
$$
\boxed{\frac{dV}{dt}=V_x\dot x+V_y\dot y=-V_x^2-V_y^2\leq0}.
$$
At the stated initial point, $V(1,-1/2)=-3/4$. Its trajectory stays in the <invariant sublevel set> $V\leq-3/4$. This set is compact because $V$ is a <coercive function>. It also excludes every point with $x=0$, where $V=y^2\geq0$; continuity therefore keeps the trajectory in its positive-$x$ component.
To justify its limit rather than merely infer it from the sketch, integrate the dissipation identity: $\int_0^\infty\|\nabla V\|^2dt$ is finite. On this compact set the gradient and its time derivative are bounded, so $\|\nabla V\|^2$ is uniformly continuous. A nonnegative uniformly continuous function with finite integral tends to zero: otherwise separated intervals of a fixed positive height and width would force an infinite integral. Every accumulation point is consequently a <critical point>. In the positive-$x$ component below this energy level the only one is $(1,-1)$. Compactness then implies convergence, and
$$
\boxed{(x(t),y(t))\longrightarrow(1,-1)\quad(t\to\infty)}.
$$
This is a <compact gradient-flow trapping criterion>; the drawn trajectory illustrates, rather than replaces, the convergence proof.
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