= Solution
Along either coordinate axis, the restriction is $h(t,0)=h(0,t)=t^6$, which is positive for $t\ne0$. Along either diagonal it is
$$
h(t,t)=h(t,-t)=2t^6-24t^4=2t^4(t^2-12),
$$
which is negative for sufficiently small nonzero $t$. Therefore every neighbourhood of the origin contains both positive and negative values, whereas $h(0,0)=0$. Consequently
$$
\boxed{(0,0)\text{ is a saddle point, neither a local minimum nor a local maximum}}.
$$
The <Hessian matrix> vanishes there, so the ordinary <second-derivative test> is inconclusive; the restrictions settle the question.
\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-origin-sections.png]
{title=Opposite signs along axes and diagonals establish the saddle at the origin}
{description=The two axes give $t^6>0$, while the two diagonals give $2t^6-24t^4<0$ near zero. The panels use different vertical scales to make both signs visible.}
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