find critical point $f(x,y) = 8 + 2y^3+2x^3 - 3xy$ $f(x,y) = 8 + 2y^3+2x^3 - 3xy$
$f_x = 6x^2 - 3y$
$f_{xx} = 12x$, $x= 0$, $y=0$
$f_y = 6y^2 -3x$
$f_{yy}=12y$
$f_{xy}=-3$
at $(0,0)$, $D = -9 < 0$
Is it one saddle point $(0,0)$?

You need to first find ALL the points where $f_x = f_y = 0$. 
$f_x = 6x^2 - 3y = 0$ gives us $y = 2x^2$ (1)
$f_y = 6y^2 - 3x = 0$ gives us $x = 2y^2$ (2)
Substitute (1) into (2) to get $x = 2(2x^2) = 8x^4$, i.e. $8x^4-x = 0$. 
This has another real root besides $x = 0$, which will give you another critical point. 
