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All the points with $0$ as a coordinate and those with $y=\pm\sqrt{1-x^2}$ are clearly global minima. After equating to $0$ the partial derivatives of the function $$f(x,y)=\log(1+\lvert xy(x^2+y^2-1)\rvert),$$ I think the stationary points are $\left(\pm\frac12,\pm\frac12\right), \left(-\frac12,\frac12\right),\left(\frac54,\frac14\right).$ I proceeded to calculate the second partial derivatives in order to construct the hessian matrices of the function at these points.

However all of these turned out to have a null determinant, if I'm not mistaken. For the other points $(x_0,y_0)$ I then studied the sign of $f(x,y)-f(x_0,y_0),$ with $(x,y)$ close to $(x_0,y_0)$. I found that $(\pm1/2,\pm1/2)$ and $(-1/2,1/2)$ are relative maxima, whereas $(5/4,1/4)$ is not an extremum point.

Did I mess up? Also, was the direct study of the sign necessary in this case?

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    It might be easier to study the function $(x,y) \mapsto |xy (x^2+y^2-1)|$, or $(x,y) \mapsto (xy (x^2+y^2-1))^2$, since the latter is differentiable. – 2017-02-13
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    @copper.hat: You're right, definitely. – 2017-02-13
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    I'm with copper.hat. $log$ is well behaved so finding the critical points of $|xy(x^2+y^2-1)|$ would be much easier. And since signs don't matter that's finding the extrema of $xy(x^2+y^2-1)$ or $yx^3 +xy^3 -1$. – 2017-02-13
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    Here is a desmos.com animation for level curves of the function when $0\le z\le 0.12$, https://www.desmos.com/calculator/lim8wceeo7 – 2017-02-13

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