$$f(x,y)=x^3+y^3-3x-12y+20$$
$\nabla f(x,y)= (3x^2-3,3y^2-12)=(0,0)$
So $3x^2=3\iff x=\pm 1$ $3y^2=12\iff y=\pm 2$
$\begin{pmatrix}
f_{xx} & f_{xy} \\
f_{yx} & f_{yy}
\end{pmatrix}=\begin{pmatrix}
6x & 0 \\
0 & 6y
\end{pmatrix}$
So at $(1,2): D>0 \text{ and }f_{xx}>0$ so it is a minimum point
$(1,-2): D<0 \text{ and }f_{xx}>0$ so it is a saddle point
$(-1,2): D<0 \text{ and }f_{xx}<0$ so it is a saddle point
$(-1,-2): D>0 \text{ and }f_{xx}<0$ so it is a maximum point
Do wy need check all the combinations of $\nabla f(x,y)=(0,0)$ like I did?
Best Answer
Your work seems fine for me. If $(x,y)$ is a (local) maximum or minimum of a differentiable function defined in an open set, then $\nabla f(x,y) = (0,0)$. So the critical points $(1,2)$, $(-1,2)$, $(1,-2)$ and $(-1,-2)$ you found are candidates for (local) maxima or minima.
If $f$ is twice differentiable, then you can use the Hessian to classify the critical points previously found - which you did pretty much correctly. But since this analysis is only local, you only have that $(1,2)$ is a local minimum and $(-1,-2)$ is a local maximum.
There are no global minimum or global maximum, since $\lim_{x \to \pm \infty} f(x,0) = \pm \infty$.