What is the stationary points (and nature of those points) of f(x,y)=12xy-x^3-y^3

the stationary points could be saddle points, maximal points and minimal points.

Critical points:
f_x = 12y - 3x^2
f_y = 12x - 2y^2.

Set these equal to 0:
12y - 3x^2 = 0 ==> y = x^2/4
12x - 3y^2 = 0 ==> 4x = y^2.

Substitution yields 4x = (x^2/4)^2 = x^4/16
==> 64x - x^4 = 0
==> x = 0, 4.

So, we have the critical points (x, y) = (0, 0), (4, 4).
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Classifying these points with the Second Derivative Test:
f_xx = -6x, f_yy = -6y, f_xy = 12
==> D = (f_xx)(f_yy) - (f_xy)^2 = 36xy - 144

Since D(0, 0) < 0, we have a saddle point at (0, 0).
Since D(4, 4) > 0, and f_xx(4, 4) < 0, we have a local maximum at (4, 4).

I hope this helps!