= Solution
{c}
On the <constraint> $x^2-z^2=1$, substitute $x^2=1+z^2$ to obtain
$$
f=1+z^2+y^2+z=\left(z+\frac12\right)^2+y^2+\frac34.
$$
Therefore the two <global minima> are
$$
\boxed{(x,y,z)=\left(\pm\frac{\sqrt5}{2},0,-\frac12\right),\qquad f_{\min}=\frac34.}
$$
Every pair $(y,z)$ is allowed on either smooth branch $x=\pm\sqrt{1+z^2}$, so these are also the only <local extrema>. There is \b[no <local maximum> or <global maximum>]: the displayed positive quadratic in $(y,z)$ has no other stationary point and is unbounded above. At the minima $F_z=-2z=1$, consistent with the hypothesis in part <3/c/i>.
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