Local integrability only guarantees that you get a distribution, but says nothing about the singular support.

To make your example a locally integrable function you need to define it almost everywhere. What is your definition of the distribution $x^{-1/2}$ for $x<0$? I take it to be $|x|^{-1/2}$ for definiteness, but other extensions to all of $R$ behave similarly. None of these distributions is regular at 0 since the definition of regularity requires ''that x $\not\in$ sing supp u if and only if there is a neighborhood U of x such that the restriction of u to U is a smooth function''. But you cannot do this in a neighborhood of zero with your function.

Thus 0 is in the singular support, and indeed, the singular support is exactly $\{0\}$. The square is a product with itself, hence the support is not disjoint. Thus the theorem is not applicable in your case.