At Jacques' cajoling, I'm turning the comments into an answer.

The two dimensional Shubert function is just the product of the one dimensional one by itself. $f(x,y) = g(x)g(y)$ where $g(x) = \sum_{j = 1}^5 j \cos( (j+1)x + j)$ is the 1 dimensional Shubert function. Observe that the local maxima are all positive, and the local minima all negative. So the minima for $f(x,y)$ occur at points $\{(x,y) : g'(x)= g'(y) = 0, f(x,y) < 0\}$. In other words, the minima of $f$ occurs at points where a maximum of $g$ is multiplied against a minimum.

Notice that there are 3 global max/min each of $g$ in the interval (-10,10), and 19 max and 20 min overall. This produces the 760 total local min of $f$ with 18 of them being global. (760 = 2 * 19 * 20, 18 = 2 * 3 * 3)

To find the extrema of the 1-d Shubert function, you evaluate its first derivative, and find that it can be simplified to a degree 6 polynomial in $\sin(x)$ and $\cos(x)$ by using the angle addition formulae. I have not evaluated the computations myself, so cannot tell you whether the expression has a closed-form algebraic solution.