= Solution
The first component of the <vector field> vanishes only when $y=0$ or $x=\pm1$. The second vanishes only when $y=1$ or $x+y=3$. Intersecting these alternatives gives exactly five <equilibria>. Their <Jacobian matrix> is
$$
M(x,y)=\begin{pmatrix}2xy&x^2-1\\y-1&x+2y-4\end{pmatrix}.
$$
At $(3,0)$,
$$
M=\begin{pmatrix}0&8\\-1&-1\end{pmatrix},\qquad\det(\lambda I-M)=\lambda^2+\lambda+8,\qquad\lambda=\frac{-1\pm i\sqrt{31}}2.
$$
The real parts are negative and the imaginary parts are nonzero, so \b[$(3,0)$ is a <stable spiral>].
At $(1,1)$ the <Jacobian matrix> is $\operatorname{diag}(2,-1)$. Its real <eigenvalues> have opposite signs, so \b[$(1,1)$ is a <saddle equilibrium>].
At $(1,2)$,
$$
M=\begin{pmatrix}4&0\\1&1\end{pmatrix},\qquad\lambda=4,1.
$$
Both <eigenvalues> are positive, so \b[$(1,2)$ is an <unstable node>].
At $(-1,1)$ the <Jacobian matrix> is $\operatorname{diag}(-2,-3)$, giving \b[a <stable node>].
At $(-1,4)$,
$$
M=\begin{pmatrix}-8&0\\3&3\end{pmatrix},\qquad\lambda=-8,3,
$$
so \b[$(-1,4)$ is a <saddle equilibrium>]. All five <equilibria> are <hyperbolic>, so these linear classifications determine their nonlinear local stability as well.
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