 Solution

Draw the even $\dot{y}=0$ nullcline through $(\pm 1,0)$ and use the arrows from part (iii). Trajectories near $(-1,0)$ spiral clockwise inward because, immediately to its right on the $x$-axis, $\dot{y}<0$. At $(1,0)$ the stable and unstable eigendirections have slopes $-4$ and $1$. Their four separatrix branches divide the portrait; the only trajectory actually reaching $(1,0)$ at finite time is the constant one, while these branches approach it as $t\to \pm \infty$. All remaining curves follow the sign field, with those in its basin spiralling into $(-1,0)$.