<s>[INST] If (3,0) is a critical point, find if f(x,y) is a local maxima, local minima or a saddle point at (3,0) So i have this function:$f(x,y)=x^3+3xy^2-4y^3+x^2-33x$
Partials derivatives in $x$,$xx$,$y$,$yy$ and $xy$
$\frac{\partial }{\partial x}=3x^2+3y^2+2x-33$
$\frac{\partial^2 }{\partial x^2}=6x+2$
$\frac{\partial }{\partial y}=6xy-12y^2=6y(x-2y)$
$\frac{\partial^2 }{\partial y^2}=6x-24y$
$\frac{\partial^2 }{\partial x\partial y}=6y$
So i have this for the point $(3,0)$
$\Delta (3,0)=36x^2-144xy+12x-48y=360$
So the point $(3,0)$ is a saddle point.
Am i correct here? If not where did i go wrong?
Thank you.
 [/INST] The solutions of the system
$$
\begin{cases}
3x^2+3y^2+2x-33=0,\\
6y(x-2y)=0
\end{cases}
$$
are:
$$(3,\,0),\;\;\left(-\frac{11}{3},\;0 \right) ,\\ \left(- \frac{4}{15} - \frac{2\cdot \sqrt{499}}{15}, \; - \frac{2}{15} - \frac{\sqrt{499}}{15}\right),\;\; \left(- \frac{4}{15} + \frac{2\cdot \sqrt{499}}{15}, \; - \frac{2}{15} + \frac{\sqrt{499}}{15}\right).$$
</s>