= Solution
Introduce the orthogonal coordinates
$$
u=\frac{x-z}{\sqrt2},
\qquad
v=\frac{x+z}{\sqrt2}.
$$
Then the constraint and objective become
$$
u^2+v^2+2y^2=1,
\qquad
\phi=\sqrt2uy.
$$
The <Lagrange multiplier> equations for a <stationary point> are
$$
\sqrt2y=2\lambda u,
\qquad
\sqrt2u=4\lambda y,
\qquad
0=2\lambda v.
$$
If $\lambda=0$, then $u=y=0$ and $v=\pm1$, giving
$$
(x,y,z)=\left(\frac1{\sqrt2},0,\frac1{\sqrt2}\right),
\quad
\left(-\frac1{\sqrt2},0,-\frac1{\sqrt2}\right).
$$
If $\lambda\ne0$, then $v=0$. The first two equations and the constraint imply
$$
\lambda=\pm\frac12,
\qquad
u=\pm\frac1{\sqrt2},
\qquad
y=\sqrt2\lambda u.
$$
Converting back to $(x,y,z)$ gives the other four points. Therefore all stationary points are
$$
\boxed{\left(\pm\frac1{\sqrt2},0,\pm\frac1{\sqrt2}\right)}
$$
with matching signs, together with
$$
\boxed{\left(\frac12,\frac12,-\frac12\right),
\left(-\frac12,-\frac12,\frac12\right),
\left(\frac12,-\frac12,-\frac12\right),
\left(-\frac12,\frac12,\frac12\right)}.
$$
The middle pair has $\phi=1/2$, the last pair has $\phi=-1/2$, and the first pair has $\phi=0$.
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