 Solution

A stationary point requires $x=z=0$ and $y(a-y^2)=0$. The Hessian matrix is diagonal: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 $\mathrm{diag}(2,-8a,2)$; both are saddle points.

For $a<0$, the origin is the only stationary point and has an indefinite Hessian matrix, so 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 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 $f$.