= Solution
The <gradient> is <orthogonal> to each regular <level set> and points towards increasing $f$. Near either <minimum>, the linear approximation is $(8\delta x,\tfrac73\delta y)$, so arrows point away from the point. Near the <maximum> at the origin it is $(-4\delta x,-\tfrac53\delta y)$, so arrows point inwards. Near either <saddle point> it is $(-2\delta x,\tfrac{24}{7}\delta y)$: arrows point towards the vertical axis horizontally and away from the saddle vertically.
On the horizontal axis the <gradient> is $(4x(x^2-1),0)$. Its arrows point left on $x<-1$, right on $-1<x<0$, left on $0<x<1$, and right on $x>1$. On the vertical axis it is
$$
\nabla f(0,y)=\left(0,\frac{y(2y^2-1)(2y^2+5)}{3+y^2}\right).
$$
Consequently its arrows point down on $y<-1/\sqrt2$, up on $-1/\sqrt2<y<0$, down on $0<y<1/\sqrt2$, and up on $y>1/\sqrt2$. These directions, and representative arrows near each <stationary point>, appear in the right-hand panel. The <gradient> itself is zero at all five <stationary points>.
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