I need to find the critical points on the boundary,inside $D$,outside $D$, and find the image of this function (constrained on $D$).
$f(x,y)=2x^2-2xy^3+3y^2$
$D=\{2x^2+3y^2\le 9\}$
Critical points non-constrained:
$f_x=4x-2y^3=0$
$f_y=-6xy^2+6y=0$ --> $6y(-xy+1)=0$
$f_y$ is equal to $0$ if $y=0$ or $x=\frac{1}{y}$, plugging it in $f_x$ brings to these critical points : $(0,0),(\pm \frac{1}{2^{\frac{1}{4}}},\pm 2^{\frac{1}{4}})$
using the second derivate test: $(0,0),(\pm \frac{1}{2^{\frac{1}{4}}},\pm 2^{\frac{1}{4}})$ seem to be saddle points (?).
Constrained Critical points (extremas):
$$ \left\{ \begin{array}{c} 4x-2y^3=4x\lambda\\ -6xy^2+6y=6y\lambda\\ 2x^2+3y^2-9=0 \end{array} \right. $$
$-6xy^2+6y=6y\lambda$ --> $6y(-x+1-\lambda)=0$ from this one I can see that it nullifies when $y=0$ or $x=1-\lambda$.
If $y=0$ the first equation gets to $4x(1-\lambda)=0$ which means that it nullifies when $x=0$ or $\lambda = 1$. Looking at the third equation tells us that $(x,y)=(0,0)$ can't be used;
Plugging $y=0$ in the third equation : possible critical points $(\pm \frac{3}{\sqrt(2)},0)$
What I can say about $(\pm \frac{3}{\sqrt(2)},0)$ ?
I think I missed some points.