= Solution
A <stationary point> requires $x=z=0$ and $y(a-y^2)=0$. The <Hessian matrix> is diagonal:
$$
H_f=\operatorname{diag}(2,4a-12y^2,2).
$$
For $a>0$, the origin has positive definite <Hessian matrix>, hence is a strict <local minimum>, with value zero. The two additional <stationary points> $(0,\pm\sqrt a,0)$ have value $a^2$ and <Hessian matrix> $\operatorname{diag}(2,-8a,2)$; \b[both are saddle points].
For $a<0$, the origin is the only <stationary point> and has an indefinite <Hessian matrix>, so \b[it is a saddle point]. For $a=0$, the <Hessian matrix> test is inconclusive, but $f(x,y,z)=x^2+z^2-y^4$ takes positive values along the $x$-axis and negative values along the $y$-axis arbitrarily near zero. Thus \b[the origin is a degenerate saddle point]. There are no <local maxima> in any case: varying $x$ or $z$ away from a <stationary point> increases $f$.
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