= Solution
Near a nonzero <stationary point> $\mathbf p$, <Taylor expansion> of the <gradient> gives $\nabla h(\mathbf p+\mathbf v)=H(\mathbf p)\mathbf v+O(|\mathbf v|^2)$. Since the <Hessian matrix> is <positive-definite matrix>[positive definite], $\mathbf v\cdot\nabla h(\mathbf p+\mathbf v)>0$ for sufficiently small nonzero $\mathbf v$. Thus the <gradient> arrows point away from each of the four <local minima>, with different strengths in the two <eigenvector> directions.
At the origin, on the $x$-axis the <gradient> is $(6x^5,0)$ and on the $y$-axis it is $(0,6y^5)$, so arrows point outward. On $y=x$ the <gradient> is $6x^3(x^2-8)(1,1)$, and on $y=-x$ it is $6x^3(x^2-8)(1,-1)$, so arrows point toward the origin for small $|x|$. Away from the axes the leading terms are $(-48xy^2,-48x^2y)$, pointing inward in both coordinate signs. The narrow outward sectors close to the axes are visible in the exact <vector field>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2-q12-gradients.png]
{title=Gradient directions near all five stationary points}
{description=Arrows show normalized <gradient> directions, not the direction of descent. They point outward near the four minima, while the degenerate origin has outward axis directions and inward diagonal directions.}
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