 Solution

The constraint implies $x^2=z^2+1$, so $x\ne 0$ and its gradient $(2x,0,-2z)$ never vanishes on the feasible set. The Lagrange multiplier equations $\nabla f=\lambda \nabla g$ giveThus $\lambda =1$, $y=0$, $z=-1/2$, and feasibility gives $x=\pm \sqrt{5}/2$. The two stationary points and their values areThey are both global minima. Indeed, on the constraint, completing the square givesEquality holds only at the two displayed points. There are no other constrained stationary points, and the objective is unbounded above as $|y|\to \infty$.