Answer
$(8,4)$ and $(-2,-1)$ are local minimum points and $(0,0)$ is a saddle point.
Work Step by Step
Let us first find the critical points of the function where $f_x$ and $f_y$ are equal to 0, assuming first and second partial derivatives are continuous.
$f_x=4x-8y$ and $f_y=-8x+4y^3-12y^2$
Let us substitute $x=2y$ from the first equation into the second equation. $f_y=-16y-12y^2+4y^3=0$
$f_y=4y(-4-3y+y^2)=4y(y-4)(y+1)=0$
$y=0$, $y=4$, and $y=-1$.
Therefore, we have the following three critical points, $(0,0)$, $(8,4)$, and $(-2,-1)$.
$f_{xx}=4$ and $f_{yy}=12y^2-24y$ and $f_{xy}=-8$
$D(0,0)=4(0)-(-8)^2<0$, which means $(0,0)$ is a saddle point.
$D(8,4)=4(192-96)=384>0$ and $f_{xx}>0$, which means the point $(8,4)$ is a local minimum.
$f(8,4)=-128$
$D(-2,-1)=4(36)>0$ and $f_{xx}>0$, which means the point $(-2,-1)$ is also a local minimum.
$f(-2,-1)=-3$
