= Solution
Use constraints $h_1=x^2/4+y^2+z^2-1$ and $h_2=x+y-z$. Both constraints and the objective are <continuously differentiable>. The feasible set is a nonempty <closed subset> of a bounded ellipsoid, hence <compact>, so the <extreme value theorem> guarantees both extrema.
The gradients $(x/2,2y,2z)$ and $(1,1,-1)$ are independent everywhere feasible. If the first were $t$ times the second, then $x=2t$, $y=t/2$, $z=-t/2$. The plane constraint would give $3t=0$, forcing the origin, which is not on the ellipsoid. Thus the <Lagrange multiplier> theorem applies at every feasible point.
Its equations are
$$
1=\lambda x/2+\mu,\qquad
2=2\lambda y+\mu,\qquad
0=2\lambda z-\mu.
$$
Here $\lambda\ne0$, since otherwise the first two equations require $\mu=1=2$. Subtract twice the first equation from the second, and use $\mu=2\lambda z$ and $z=x+y$; the result is $0=-3\lambda x$, so $x=0$. Then $z=y$ and the ellipsoid gives $y=\pm1/\sqrt2$. The only candidates are therefore $(0,\pm1/\sqrt2,\pm1/\sqrt2)$, with matching signs. Consequently
$$
\boxed{\min g=-\sqrt2,\qquad\max g=\sqrt2}.
$$
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