<s>[INST] 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
 [/INST] $$f_x=3x^2-12y$$
$$f_y=24y^2-12x$$
Now we let $\nabla f(x,y)=(f_x,f_y)=0$ which gives us the following:
$$
\left\{ 
\begin{array}{c}
f_x=3x^2-12y=0 \\ 
f_y=24y^2-12x=0
\end{array}
\right. 
$$
Solving it we have
$$
\left\{ 
\begin{array}{c}
x=2y^2 \\ 
12y^4-12y=12y(y^3-1)=0
\end{array}
\right. 
$$
Where $y=0\lor y=1$ and respectively $x=0\lor x=2$. Can you continue now?
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