Find the local extrema and saddle points of f(x,y) = 2x^3 + 4y^3 + 3x^2 - 2y^2 +1.

For the critical points:
f_x = 6x^2 + 6x
f_y = 12y^2 - 4y.

Setting these equal to 0 yields 4 critical points
(x, y) = (0, 0), (-1, 0), (0, -1/3), (-1, -1/3).
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Classify these with the second derivative test.
f_xx = 12x + 6, f_y = 24y - 4, f_xy = 0
==> D = (12x + 6)(24y - 4) - 0^2 = (12x + 6)(24y - 4).
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Since D(0, 0) = -24 < 0, (0, 0) is a saddle point.
Since D(-1, 0) = 24 > 0 and f_xx(-1, 0) = -6 < 0, we have a local maximum at (-1, 0).
Since D(0, -1/3) = -72 < 0, we have a saddle point at (0, -1/3).
Since D(-1, -1/3) = 72 > 0, and f_xx (-1, -1/3) = -6 < 0, we have a local maximum at (-1, -1/3).

I hope this helps!