Find all critical points of $f(x,y) = x^3 - 12xy + 8y^3$ and state maximum, minimum, or saddle points. 
Find all critical points of $f(x,y) = x^3 - 12xy  + 8y^3$ and state whether the function has a relative minimum, relative maximum, or a saddle at the critical points.

So I have:
$f_x = 3x^2 -12 y$
$f_y = -12x + 24y^2$

$f_{xx} = 6x$
$f_{yy} = 48y$
$f_{xy} = -12$

I found that my critical points were: $(0,0)$ and $(2,1)$, but I still need to classify them.
My question is, how do I check if $f_{xx}$ is positive or negative? This seems like a silly question (I'm sure it is), but do I plug in $(0,0)$ for $f_{xx}$? Wouldn't that then be $6\cdot 0 > 0$? That seems wrong.
 A: Given:
$$\tag 1 f(x,y) = x^3 - 12xy  + 8y^3$$
Find and classify all critical points.
The critical points are: $(0,0)$ and $(2,1)$.
The partial derivatives are:


*

*$f_x = 3x^2 -12 y$

*$f_y = -12x + 24y^2$

*$f_{xx} = 6x$

*$f_{yy} = 48y$

*$f_{xy} = f_{yx} = -12$


The Hessian determinant is given by:
$$\det(H) = \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{vmatrix} \ $$
If you are using the Hessian, there are four conditions you need to test:


*

*$(01)$ $f_{xx} \gt 0$ and $\det H \gt 0 \rightarrow$ local minimum

*$(02)$ $f_{xx} \lt 0$ and $\det H \gt 0 \rightarrow$ local maximum

*$(03)$ $\det(H) \lt 0 \rightarrow$ saddle point

*$(04)$ $\det(H) = 0 \rightarrow$ no statement can be made using this approach


For $(0,0)$, condition $(4)$ tells us that $\det H = 0$, so nothing can be said about this critical point, neither a min or max. Of course, we could have also looked at $$\displaystyle g(x,y) = f_{xx}(x, y)f_{yy}(x,y)-[f_{xy}(x,y)]^2.$$
Since $g(x,y) \lt 0$, this is not an extremum. 
For $(2,1)$, we have: $f_{2,1} = 12 > 0$ and $\det H = 432 >0 \rightarrow$ a local minimum. The value of $f(x,y)$ at this minimum is $f(2,1) = -8$.
Plots are given by:


