= Solution
Use $g=x^2+y^2+z^2$. The <Lagrange multiplier> equations are
$$
y=2\lambda x,\qquad x+z=2\lambda y,\qquad y=2\lambda z,
\qquad x^2+y^2+z^2=1.
$$
Taking the <dot product> with $(x,y,z)^T$ gives $2T=2\lambda$, so $\lambda=T$.
If $\lambda\ne0$, the first and third equations give $x=z$. The middle equation then gives $x=\lambda y$, so $2\lambda^2=1$. The norm constraint gives $x^2=z^2=1/4$ and $y^2=1/2$. Thus the hottest points and their temperature are
$$
\boxed{(x,y,z)=\left(\frac\varepsilon2,\frac\varepsilon{\sqrt2},\frac\varepsilon2\right),\quad
T_{\max}=\frac1{\sqrt2},\quad\varepsilon\in\{-1,1\}}.
$$
The coldest points are
$$
\boxed{(x,y,z)=\left(\frac\varepsilon2,-\frac\varepsilon{\sqrt2},\frac\varepsilon2\right),\quad
T_{\min}=-\frac1{\sqrt2},\quad\varepsilon\in\{-1,1\}}.
$$
For $\lambda=0$, the equations give $y=0$, $z=-x$, and hence the other two constrained <stationary points> are
$$
\boxed{(x,y,z)=\left(\frac\varepsilon{\sqrt2},0,-\frac\varepsilon{\sqrt2}\right),\qquad T=0}.
$$
They are saddle points of the restriction to the <sphere>: with $u=(x+z)/\sqrt2$ and $v=y$ as local tangent coordinates there, $T=\sqrt2uv$ changes sign in every neighbourhood. The surface is compact and smooth, so the listed extreme temperatures are global. Here “zero gradient on the surface” means zero <surface gradient>; the ambient <gradient> at the hottest and coldest points is normal to the <sphere>, rather than zero.
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