[tex]f(x,y)=2y^3+12x^2-24xy[/tex]
Find the first derivatives:
[tex]f'_x=24x-24y \\ f'_y=6y^2-24x[/tex].
Solve the system [tex] \left \{ {{f'_x=0} \atop {f'_y=0}} \right. [/tex]:
[tex] \left \{ {{24x-24y=0} \atop {6y^2-24x=0}} \right. \rightarrow \left \{ {{x=y} \atop {6y^2-24y=0}} [/tex]. The second equation has solutions [tex]y_1=0, \\ y_2=4[/tex] and then [tex]x_1=0, \\ x_2=4[/tex] and you have two points [tex]A_1(0,0), \\ A_2(4,4)[/tex].
Find the first derivatives:
[tex]f^{''}_{xx}=24 \\ f^{''}_{xy}=f^{''}_{yx}=-24 \\ f^{''}_{yy}=12y[/tex] and calculate [tex]\Delta=\left| \left[\begin{array}{cc}24&-24\\-24&12y\end{array}\right]\right |=24\cdot 12y-(-24)^2=288y-576[/tex].
Since [tex]\Delta(A_1)=-576 [/tex] and [tex]f^{''}_{xx}\ \textgreater \ 0[/tex], [tex]A_1[/tex] is a point of maximum and [tex]f(0,0)=0[/tex].
Since [tex]\Delta(A_2)=576 [/tex] and [tex]f^{''}_{xx}\ \textgreater \ 0[/tex], [tex]A_1[/tex] is a point of minimum and [tex]f(4,4)=-64[/tex].
