= Solution
Draw the downward-opening <nullcline> $y=x(2-x)$, meeting the axes at $x=0,2$, and the line $y=kx$. Their positive intersection is $P=(2-k,k(2-k))$. Use the sign rules
$$
\dot x>0\iff y<x(2-x),\qquad \dot y>0\iff y>kx.
$$
Above both curves the arrows point northwest; below both they point southeast. For $0<x<2-k$, the parabola lies above the line, so arrows between them point northeast. For $2-k<x<2$, their order reverses, so arrows between them point southwest. For $x>2$, $\dot x<0$ throughout the nonnegative quadrant, with the vertical direction still determined by the line. Mark the origin as an unstable node and $P$ as a saddle, and trace sample trajectories consistent with those directions.
\Image[/past-exam-of-the-differential-equations-1-course-of-the-university-of-oxford/2017-phase-plane.png]
{title=<Nullclines> and sample trajectories for the representative value $k=1$}
{description=The displayed vector field is restricted to the nonnegative quadrant. The origin is an unstable node and the interior equilibrium is a saddle.}
The quadrant is \b[not positively invariant]: on $x=0,y>0$, $\dot x=-y<0$; on $y=0,x>0$, $\dot y=-kx<0$. Trajectories reaching either axis there leave the plotted region. The <phase portrait> describes the portions inside the quadrant, not a claim that all forward trajectories remain inside it.
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