= Solution
The constraint implies $x^2=z^2+1$, so $x\ne0$ and its <gradient> $(2x,0,-2z)$ never vanishes on the feasible set. The <Lagrange multiplier> equations $\nabla f=\lambda\nabla g$ give
$$
2x=2\lambda x,\qquad2y=0,\qquad1=-2\lambda z.
$$
Thus $\lambda=1$, $y=0$, $z=-1/2$, and feasibility gives $x=\pm\sqrt5/2$. The two stationary points and their values are
$$
\boxed{\left(\frac{\sqrt5}{2},0,-\frac12\right),\
\left(-\frac{\sqrt5}{2},0,-\frac12\right),\qquad f=\frac34.}
$$
They are both <global minima>. Indeed, on the constraint, <completing the square> gives
$$
f=1+z^2+y^2+z=\left(z+\frac12\right)^2+y^2+\frac34.
$$
Equality holds only at the two displayed points. There are no other constrained stationary points, and the objective is unbounded above as $|y|\to\infty$.
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