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I am having trouble finding the partial derivative. And clues or hints regarding said problem and how to find saddle points/local maxima is appreciated.

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    Note that if you set the derivatives to zero and solve, you will get the only stationary point $x=y=-\sqrt[3]{2}$. If you substitute this into the second derivative, you will get the matrix $\pmatrix{2 & 1 \\ 1 & 2}$, which is positive definite, hence this is the only local minimum.2012-11-24

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