= Solution
The curve is symmetric in both coordinate axes and under interchange of $x$ and $z$. Its shape, including its eight <inflection points>, is:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1-curve-and-inflections.png]
{title=A plane curve and its <inflection points>}
Writing $r^2=x^2+y^2$, the <surface of revolution> has equation
$$
\boxed{(x^2+y^2-1)^2+(z^2-1)^2=5}.
$$
Let
$$
F(x,y,z)=(x^2+y^2-1)^2+(z^2-1)^2-5.
$$
Its <gradient> is
$$
\nabla F=4\bigl(x(r^2-1),y(r^2-1),z(z^2-1)\bigr).
$$
If this vanished, then $r^2\in\{0,1\}$ and $z^2\in\{0,1\}$, but at such a point $(r^2-1)^2+(z^2-1)^2\le2$, not five. Hence zero is a <regular value>, and the <regular level set theorem> shows that $S=F^{-1}(0)$ is a smooth <embedded submanifold>[embedded surface].
For a surface of revolution, the two <principal curvatures> are the curvature of the meridian and the normal curvature of a parallel. Along $r=1$,
$$
F_r=4r(r^2-1)=0,
$$
so the normal is vertical in the meridian plane and the parallel principal curvature vanishes. Therefore the <Gaussian curvature>, the product of the principal curvatures, is zero there.
These are not the only zero-curvature points. The meridian itself has inflection points, at which its principal curvature vanishes. Setting $u=r^2$ and $v=z^2$, their positive squared coordinates satisfy
$$
(u-1)^2+(v-1)^2=5,
$$
$$
(3u-1)v(v-1)^2+(3v-1)u(u-1)^2=0.
$$
Besides interchanging $u$ and $v$, the solution is approximately
$$
(u,v)=(0.304199,3.125056).
$$
Each corresponding meridian inflection point sweeps out another circle on $S$, so $r=1$ does not exhaust the zero set of the Gaussian curvature.
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