Extrema of a surface $z=f(x;y)$ when $det(H)=0$

Question

I'm given the following problem:

$\text{Examine}\ z=f(x;y)=x^4+y^4+18xy-9x^2-9y^2+1\text{ for extrema and saddle points.}$

It is trivial to find $\nabla f=(4x^3+18y-18x; 4y^3+18x-18y)$ and the critical points are $P_0(0,0), P_1(3,-3)\text{ and }P_2(-3,3)$.

Also the determinant of the Hessian matrix is easily found (details not included): $$\text{det}(H)=72(2x^2y^2-3x^2-3y^2)$$

Evaluating $\text{det}(H)$ for $P_1$ and $P_2$ gives positive values so there are extrema in these points. Furthermore $f''_{xx}>0$ so these are both minima.

For $P_0$ the determinant is $0$. In our class it was said that in these cases "further investigation is needed", but noone provided me with any information about this "further investigation". In internet I also found nothing.

I plotted the graph of $z=x^4+y^4+18xy-9x^2-9y^2+1$ in GeoGebra and it turned out that in $P_0$ there is a saddle point. But I want to find out how can one analytically determine what to further do when $\text{det}(H)=0$

When in a single-variable calculus $f'(x)=f''(x)=0$, I keep differentiating $f$ until I get a non-zero derivative. And if the first non-zero derivative is of odd order (i.e. $f^{(3)}, f^{(5)}$ and so on), I know the function has an inflection point; when it is of even order, there is an extremum, whose kind depends on the sign of this non-zero derivative.

And so I'm stuck with this problem. I'm not explicitly asked to find what is there in $P_0(0,0)$, but I want to learn how to tackle such problems. Any help will be appreciated. Thanks in advance!

$\textbf{Edit:}$ The answer by @Robert Z is very helpful, but I find it a bit of a guess-and-check method. And if at $P_0(0,0)$ there was an extremum, it wouldn't work out because we can't check all lines, passing through $P_0$. Any suggestions for the case where there would be an extremum?

Answer 1

In order to find the nature of the stationary point $(0,0)$ you may restrict the function along two suitable lines passing through the origin. Noting that $$f(x,y)=x^4+y^4+18xy-9x^2-9y^2+1=x^4+y^4-9(x-y)^2+1$$ we find that along the line $y=x$, $$h(x):=f(x,x)=2x^4+1\geq 1=h(0)$$ has a local strict minimum at $x=0$.

On the other hand, along the line $y=-x$, for $|x|<3$, $$g(x):=f(x,-x)=2x^4-36x^2+1=2x^2(x^2-18)+1\leq 1=g(0)$$ has a strict local maximum at $x=0$.

Thus we may conclude that the stationary point $(0,0)$ is a saddle point.