= Solution
Because the constraint determines
$$
x=b^2-a^2y^2-z^2,
$$
the constrained function is
$$
\Phi(y,z)=yz(b^2-a^2y^2-z^2).
$$
Its <stationary points> satisfy
$$
\frac{\partial\Phi}{\partial y}
=z(b^2-3a^2y^2-z^2)=0,
\qquad
\frac{\partial\Phi}{\partial z}
=y(b^2-a^2y^2-3z^2)=0.
$$
When $yz=0$, these equations give
$$
(x,y,z)=(b^2,0,0),\quad
(0,\pm b/a,0),\quad
(0,0,\pm b).
$$
When $yz\ne0$, subtracting the two bracketed equations gives $z^2=a^2y^2$, and substitution then gives
$$
x=\frac{b^2}{2},
\qquad
y=\pm\frac b{2a},
\qquad
z=\pm\frac b2,
$$
with the two signs chosen independently. These are exactly the constrained stationary points; equivalently they follow from the <Lagrange multiplier> equations.
Under the additional restriction $x\ge0$, the constraint gives the compact elliptical disc $a^2y^2+z^2\le b^2$. On its boundary $x=0$, the objective is zero. At the four nonzero interior stationary points,
$$
\phi=xyz=\pm\frac{b^4}{8a},
$$
with the sign determined by $yz$. Therefore
$$
\boxed{\max\phi=\frac{b^4}{8a},
\qquad
\min\phi=-\frac{b^4}{8a}}.
$$
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