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1) How do I take the second derivative of ky 2 -2y 3 -kx 3 +2kxy, with a critical point at (0,0). 2) Why is

1) How do I take the second derivative of ky2-2y3-kx3+2kxy, with a critical point at (0,0).

2) Why is there a "saddlebag" at the critical point for every non-zero value of k?

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