Find local and global extrema for $f(x,y) = y^4 -3xy^2 +x^3$ above you find a function and some questions I have to answer. I'll give you a more or less detailed input of what I did.
I'll be glad if you could help me with the questions I inserted with "->".
Given is $f(x,y) = y^4 -3xy^2 +x^3$
First I have to determine the local and global extrema and the saddle points.
Then I have to find the extrema on $ D = \{ (x,y) : |x| \le 5/2, |y| \le 2 \}$
Short: 


*

*$grad f (x,y) = (-3y^2+3x^2, 4y^3-6xy)^t$

*$grad f (x,y) = 0 \leftrightarrow x^2 = y^2 \wedge y^2/x = 3/2 $

*So possible extrema are at $ (3/2, 3/2), (3/2, -3/2)$ and $(0,0)$.

*The determinant of the hesse matrix is: $Det H_f (0,0) = 0$, $Det H_f (3/2, 3/2) = 81$ and $Det H_f (3/2, -3/2) = 81$. 

*I conclude: at the point $(0,0)$ I cannot tell wether there is a minima or a maxima. At $(3/2, 3/2)$ and $(3/2, -3/2)$ I have local minima. I have no global maxima or minima, since $f(x,y) \to \pm \infty$ for $(-\infty, 0)$ and $(0, \infty$ (for example)
-> I wonder, if I truly cannot say anything about the extrema at $(0,0)$ ? Is it always like this: If determinant of hesse matrix at $(x_0, y_0)$ is $(0,0)$ than you cannot say if it's indefinite or positiv/negativ definite?


*for the second part I simply put $f(5/2, y)$ and $f(-5/2, y)$ as well as $f(x, -2)$ and $f(x, 2)$. Then I look at the first derivates and see possible maxima/minima -> I put those points in the regarding function and see, that $f(-2,2) = 32$ is Maximum - $f(-5/2, 0)$ is Minimum.


-> I wonder, if there is a better (meaning: mathematical cooler, more elegant) way to find this out? I thought about cutting my original function with a certain layer?!
Thank you very much for any help :)
 A: Well, one approach you can take in this particular situation is to observe that $f$ can be rewritten as $$\left(y^2\right)^2-3xy^2+x^3,$$ which is a quadratic in $y^2$. Hence, for any fixed $x,$ one can calculate directly which (if any) values of $y$ will minimize $f$. Namely, one proceeds with one of our most basic (and surprisingly useful) results: the difference of squares formula!
Readily, $$\begin{align}f(x,y)&=\left(y^2-3x\right)y^2+x^3\\&=\left(y^2-\frac32x-\frac32x\right)\left(y^2-\frac32x+\frac32x\right)+x^3.\end{align}$$ Applying the difference of squares formula yields $$f(x,y)=\left(y^2-\frac32x\right)^2-\frac94x^2+x^3.\tag{$\star$}$$ Clearly, then, for fixed $x,$ we see that $f(x,y)$ is minimized when $y^2=\frac32x,$ and in such a case takes on the value $$x^3-\frac94x^2=x^2\left(x-\frac94\right).\tag{$\heartsuit$}$$ We can then find $x$ that minimize $(\heartsuit)$ locally for $|x|<\frac52.$ The only such value is $\frac32,$ and so we obtain the same minima through this method.
It seems that we ought to be able to conclude that $f(x,y)$ has no local maxima in this way, but I can't see how we might do so rigorously, offhand, without simply resorting to more or less the same techniques that you employed. However, we can note that $x=0$ locally maximizes $(\heartsuit),$ which suggests that there probably isn't a local maximum at the origin. Indeed, while $f(0,y)=y^4$ is minimized at $y=0,$ we see that $f(x,0)=x^3$ has an inflection point at $x=0$. Hence, we can rule out the origin as a possible local extremum. It might even be possible to show that there is a saddle point at the origin, but again, I'm not seeing how to do so offhand with this approach.
