= Solution
The <equilibrium> equations are $x(1+x-2y)=0$ and $(x-1)y=0$. If $x=0$, the second equation gives $y=0$. Otherwise either $y=0$, giving $x=-1$, or $x=1$, giving $y=1$. Thus there are exactly three equilibria. The Jacobian is
$$
J(x,y)=\begin{pmatrix}1+2x-2y&-2x\\y&x-1\end{pmatrix}.
$$
At $(0,0)$ it is $\operatorname{diag}(1,-1)$: \b[a <saddle equilibrium>[saddle]]. At $(-1,0)$ it is $\begin{pmatrix}-1&2\\0&-2\end{pmatrix}$, with <eigenvalues> $-1,-2$: \b[a <stable node>]. At $(1,1)$ it is $\begin{pmatrix}1&-2\\1&0\end{pmatrix}$, with <characteristic polynomial> $\lambda^2-\lambda+2$ and <eigenvalues> $(1\pm i\sqrt7)/2$: \b[an <unstable spiral>]. All three are hyperbolic, so the nonlinear local classifications follow from <linearization>.
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