= Solution
Complete the square:
$$
h=(x^2+2y)^2-6.
$$
The <minimum locus of a squared function> occurs where its residual vanishes. Hence the minimum is $-6$ on $y=-x^2/2$, provided these points lie in the ellipse. Setting $t=x^2$ in the constraint gives $t+t^2/2\leq9$, or $0\leq t\leq\sqrt{19}-1$. Thus
$$
\boxed{\min_R h=-6,\quad\text{at }\left\{(x,-x^2/2):|x|\leq\sqrt{\sqrt{19}-1}\right\}.}
$$
For the maximum, let $q=x^2+2y$. The ellipse gives
$$
q\leq9-2y^2+2y=\frac{19}{2}-2\left(y-\frac12\right)^2\leq\frac{19}{2}.
$$
Also $q\geq2y\geq-3\sqrt2$, whose absolute value is less than $19/2$. Therefore $q^2\leq(19/2)^2$. Equality is attained at $y=1/2$, $x^2=17/2$, giving
$$
\boxed{\max_R h=\frac{337}{4}=84\frac14,\quad\text{at }(x,y)=\left(\pm\sqrt{17/2},\frac12\right).}
$$
These attained bounds are global, not merely stationary-point tests. The minimizing locus is a parabola segment.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/continuous-mathematics-quartic-extrema.png]
{title=Global extrema of the squared-residual quartic on an ellipse}
{description=Contours show h inside the feasible ellipse. The blue parabola segment is the entire minimum set h = -6; the two marked boundary points attain the maximum 337/4.}
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