Relative max/min in multivariable calculus? Please help!

Find local maximum and minimum values and saddle point(s) for the function:
f(x,y) = x^(3)*y+12x(2)-8y

My calculations:
df/dx = 3xy+24x
df/dy = x^(3)-8
d^(2)f/dx^(2) = 3y+24
d^(2)f/dy^(2) = 0
d^(2)f/dxy = 3

I got critical points of:
(0,0)
(2,0)
(0,-8)

I don't know if that's right. For example, df/dy=0 when x=2. The y variable isn't present in this equation so how do I know the critical point when x=2? When I plug it into df/dx = 0 I get y=-8. So I guess the point is actually (2,-8).

What about when solving df/dx? I get df/dx=0 when x=0 or y=-8. I can't plug either of these values into df/dy to get anything. So is the point (0,-8)?

Please help me with this. I know it's really simple but I'm having trouble piecing it all together. Some of the example problems showed how do it by solving one partial derivative equation so that one variable is expressed as a function of the other, but I don't see that being applicable here. Is that right?

From f_x = 3x^2 y + 24x and f_y = x^3 - 8:

Setting f_y = 0 yields x^3 - 8 = 0 ==> x = 2 (ignore the imaginary roots).
Setting f_x = 0 with x = 2 yields 12y + 48 = 0 ==> y = -2.

So, we have exactly one critical point (x, y) = (2, -2).
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Classifying this point:
f_xx = 6xy + 24, f_xy = 3x^2, f_yy = 0.
==> D = (f_xx)(f_yy) - (f_xy)^2 = -9x^4.

Since D(2, -2) < 0, it is a saddle point.

I hope this helps!