= Solution
The <fixed points> are $(0,0)$ and $(\pm1,1)$. The <Jacobian matrix> is
$$
J(x,y)=\begin{pmatrix}1-y&-x\\2x&-1\end{pmatrix}.
$$
At the origin its <eigenvalues> are $1,-1$, so it is a saddle with stable direction the y-axis. At either other <fixed point>, the characteristic <polynomial> is $\lambda^2+\lambda+2$, giving <eigenvalues> $(-1\pm i\sqrt7)/2$. Thus both are stable foci. The <nullclines> are $x=0$, $y=1$ and $y=x^2$. In the right half-plane $x$ grows below $y=1$ and decreases above it; in the left half-plane these horizontal arrows reverse. Vertically, $y$ increases below $y=x^2$ and decreases above it.
A global argument determines the requested <omega-limit sets>, rather than guessing from the local <phase portrait>. For $x\ne0$, use the <quadratic-logarithmic Lyapunov function for a feedback system>
$$
L=x^2-1-\log(x^2)+(y-1)^2,\qquad
\dot L=-2(y-1)^2.
$$
Its sublevel sets are <compact> and bounded away from $x=0$, so trajectories starting in either half-plane exist for all positive time and stay in such a set. Since $L$ decreases, it tends to a limit, and is constant on the <omega-limit set>. Every point of that set therefore has $y=1$. Invariance then requires $\dot y=x^2-1=0$, so only the corresponding focus can occur. Also $x(t)=x(0)\exp(\int_0^t(1-y(s))\,ds)$ preserves its sign.
Consequently
$$
\boxed{\omega(2,-1)=\{(1,1)\},\qquad
\omega(x,y)=\{(0,0)\}\ \Longleftrightarrow\ x=0}.
$$
On $x=0$, the exact solution is $y(t)=y(0)e^{-t}$. The <phase portrait> below shows this stable axis, the <nullclines> and the spiral attraction in each half-plane.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/ii/paper-4-phase-plane.png]
{title=Phase portrait with a saddle at the origin and stable foci at plus or minus one, one}
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