= Solution
Write the vector field as
$$
\dot x=2x(4-x-y^2),\qquad \dot y=y(x-1).
$$
The <equilibrium points of a dynamical system> in the closed first quadrant are
$$
\boxed{(0,0),\quad(4,0),\quad(1,\sqrt3)}.
$$
The <Jacobian matrix> is
$$
J(x,y)=
\begin{pmatrix}
8-4x-2y^2&-4xy\\
y&x-1
\end{pmatrix}.
$$
At $(0,0)$ its eigenvalues are $8,-1$, so this is a <saddle equilibrium>. Its unstable direction is the positive $x$-axis and its stable direction is the positive $y$-axis; to first order, nearby trajectories satisfy $\dot x=8x$, $\dot y=-y$.
At $(4,0)$ the eigenvalues are $-8,3$, so it is also a saddle. The $x$-axis is the stable direction, while trajectories entering the quadrant in the $y$ direction move away.
At $(1,\sqrt3)$,
$$
J=\begin{pmatrix}-2&-4\sqrt3\\ \sqrt3&0\end{pmatrix},
$$
whose eigenvalues are
$$
-1\mathbin{\pm}i\sqrt{11}.
$$
It is therefore a <stable spiral>. A point immediately to its right moves upward, so nearby trajectories spiral counterclockwise into the equilibrium. These eigendirections and the inward spiral give the requested local phase-portrait sketches.
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