Calc 3 help! Critical Points!!!

Find the four critical points of f(x, y) = 3x−x^3−xy^2, and classify each critical point as a local maximum,
local minimum, or a saddle point.

Please help me!!!

Love, Sara

First, set the first partial derivatives equal to 0 to find the critical points.

f_x = 3 - 3x^2 - y^2
f_y = -2xy.

Set these equal to 0; the second equation yields x = 0 or y = 0.
(i) If x = 0, then 3 - 0 - y^2 = 0 ==> y = ±√3.
(ii) If y = 0, then 3 - 3x^2 - 0 = 0 ==> x = ± 1.

Hence, we have the four critical points
(x, y) = (0, ±√3), (± 1, 0).
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Next, we classify these with the Second Derivative Test.
f_xx = -6x, f_yy = -2x, f_xy = -2y
==> D = (f_xx)(f_yy) - (f_xy)^2 = 12x^2 - 4y^2.

Since D(0, ±√3) < 0, we have saddle points at (0, ±√3).
Since D(1, 0) > 0 and f_xx(1, 0) < 0, we have a local maximum at (1, 0).
Since D(-1, 0) > 0 and f_xx(-1,0) > 0, we have a local minimum at (1, 0).

I hope this helps!

Type f(x, y) = 3x−x^3−xy^2 on www.wolframalpha.com

I need critical points too. 5 points until I can ask a question :X