Calculus Find Min and Saddle Point

Find the local maximum and minimum values and saddle point(s) of the function. If you have three dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.
f(x, y) = x^3 − 12xy + 8y^3

Find the minimum:
f(______________, ____________) = ________________ smallest x value
f(______________, ____________) = ________________ largest x value

Find saddle point
(______________, ____________) smallest x value

Please show all work! Thanks for your help!!

Critical points.
f_x = 3x^2 - 12y, f_y = -12x + 24y^2.

Setting these equal to 0:
3x^2 - 12y = 0, -12x + 24y^2 = 0
==> x^2 = 4y, x = 2y^2
==> 4y^4 = 4y ==> y = 0 or 1.
Hence, we have the critical points (x, y) = (0, 0), (2, 1)

Classification via second derivative test.
f_xx = 6x, f_yy = 48y, f_xy = -12
==> D = (f_xx)(f_yy) - (f_xy)^2 = 288xy - 144.

Since D(0, 0) < 0, we have a saddle point at (0, 0).
Since D(2, 1) > 0 and f_xx(2, 1) > 0, we have a local minimum at (2, 1) and f(2, 1) = 4.

I hope this helps!

f(x,y) = x³ − 12xy + 3y³

Find the critical points by taking the derivative and setting it equal to 0.

∂f/∂x = 3x² − 12y = 0 → [y = x²/4]
∂f/∂y = -12x + 24y² = 0 → [x = 2y²]

Cross-reference the functions:
y = (2y²)²/4
y = y^4
y^4 − y = 0
y(y³ − 1) = 0
y(y − 1)(y² + y + 1) = 0
y = 0, 1

x = 2y²
x = 0, 2
Critical points:
{(0,0) , (2,1)}

Second Derivatives:
∂²f/∂x² = 6x
∂²f/[∂x∂y] = ∂²f/[∂y∂x] = -12
∂²f/∂y² = 48y

Hessian Matrix:
[∂²f/∂x² ........ ∂²f/(∂x∂y)] == [ 6x ... -12 ]
[∂²f/(∂y∂x) ....... ∂²f/∂y² ] == [ -12 .. 48y ]

M = 288xy − 144

M(0,0) < 0 → Saddle Point
M(2,1) > 0 and ∂²f/∂x² > 0 → Minimum

Not sure what the other minimum is.