Max and Min$ f(x,y)=x^3-12xy+8y^3$ First I solved $f_y=0$ then plugged in my variable into $f_x$ to get an output and then plugged that output back into $f_y$ to get a point and did this again for $f_x$ I think I screwed up on the $(-7,\sqrt{3})$
$f_y=-12x+24y^2$
$-12x+24y^2=0$
$-12x=24y^2$
$x=2y^2$
$f_x(2y^2,y)=3(2y^2)^2-12y$
$3(4y^4)-12y=0$
$12y^4 - 36y=0$
$12y(y^3-3=0) \to y=0$ or $y=\sqrt{3}$
$f_y(x,0)=-12x=0 \to (0,0)$
$f_y(x,\sqrt{3}) = -12x+24(3)$
$f_y(x.sqrt{3})= -12x-84=0$
$f_y(x,\sqrt{3})=-12x=84$
$x=-7 \to (-7,\sqrt{3})$
$f_y=-12x+24y^2=0$
$-12x=24y^2$
$x=2y^2$
$f_x(2y^2,y)=3(2y^2)^2-12y=0$
$12y^4-12y=0$
$12y(y^3-1)=0$
$y=0$ or $y=1$
$f_y(x,o)= -12x=0 \to x=0$
$f_y(x,1)=-12x+24=0$
$f_y(x,1)=x=2$
Critical points: $(2,-1),(0,0),(-7,\sqrt{3})$
$D=(6x)(48y)-(-12)^2$
$D(0,0)=-144 so (0,0) is a saddle point
$D(,2,-1)= Min 
D(-7,\sqrt{3})= Saddle point

$\hskip 2.1in$
Given function $$f(x,y) = x^3 - 12xy + 8y^3$$ $$ \Rightarrow f_x = 3x^2 - 12y, f_y = 24y^2 -12x = 0 $$
Which on solving gives two stationary points $(2,1)$ and $(0,0)$.
Now, consider hessian matrix $$H = \begin{bmatrix} 6x & -12 \\ -12 & 48y \end{bmatrix} $$
For $(2,1)$, $|H| > 0$ and $f_{xx} ( 2,1) > 0$, implying that there is a minima at that point.
For $(0,0)$, $|H| <0$, implying there is a saddle point at $(0,0)$
