Calculus - partial derivatives, relative extreme values

Please help me find the relative extreme values of this function:

f(x,y) = x^3 -2xy + 4y

Critical points:

f_x = 3x^2 - 2y
f_y = -2x + 4.

Set these equal to 0 to see that x = 2 and thus y = 6.
So, the critical point is (2, 6).
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Second Derivative Test:

f_xx = 6x, f_yy = 0, f_xy = -2
==> D = (f_xx)(f_yy) - (f_xy)^2 = 0 - 4 < 0.

Hence, (2, 6) is a saddle point.

I hope this helps!