= Solution
A <smooth surface> in $\mathbb R^3$ is a subset locally parametrized by a smooth map of two variables whose derivative has rank two and which is a homeomorphism onto its image.
Writing $\rho=\sqrt{x^2+y^2}$, the first surface equation becomes
$$
(\rho-2\sqrt2)^2+z^2=1.
$$
It is the <torus> of major radius $R=2\sqrt2$ and minor radius $r=1$, with smooth parametrization
$$
\boxed{X(u,v)=((2\sqrt2+\cos v)\cos u,
(2\sqrt2+\cos v)\sin u,\sin v)},
\qquad u,v\in\mathbb R/(2\pi\mathbb Z).
$$
Since $R>r$, this is an embedding. The <Gaussian curvature of a torus> is
$$
\boxed{K(u,v)=\frac{\cos v}{2\sqrt2+\cos v}}.
$$
For the second surface, use the <orthogonal transformation>
$$
X=\frac{x+z}{\sqrt2},
\qquad Y=y,
\qquad Z=\frac{z-x}{\sqrt2}.
$$
Its equation becomes the same torus equation $(\sqrt{X^2+Y^2}-2\sqrt2)^2+Z^2=1$. Orthogonal transformations are Euclidean <isometries> and preserve <Gaussian curvature>. Curvature vanishes where $\cos v=0$, equivalently $Z=\pm1$. In the original coordinates the zero-curvature points are therefore exactly
$$
\boxed{z-x=\pm\sqrt2,
\qquad (x+z)^2+2y^2=16}.
$$
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