The derivative $\frac{\mathrm d}{\mathrm dx} x^x=x^x\left(\ln x+1\right)$ is problematic for $x<0$

Not at all suspicious. You can't differentiate a function at an isolate point of the domain. So even if you extend the domain of $x^x$ to the negative integers, you cannot differentiate it at these points: how do you do the limit?

One could define $x^x$ for negative rational values of $x$ having odd denominators.

The set $W=\{a/b: a,b\in\mathbb{Z}, a<0, b>0, b\text{ odd}\}$ is even dense in $(-\infty,0)$, so it could be a good candidate for doing limits over it.

There is a problem, though: consider $-1/3$. In every neighborhood of $-1/3$ there are points $x_0$ in $W$ having even numerator and also points $x_1$ in $W$ having odd numerator. The value of $x^x$ at $x_0$ is positive, the value of $x^x$ at $x_1$ is negative. Therefore the function is not continuous at $-1/3$.

Hence, differentiability is out of question.

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If you consider $x^x$ over the complex numbers, you have to choose a branch cut for the complex logarithm. Then the function is well defined and even analytic: $x^x=\exp(x\log x)$. Of course, in order to consider the derivative at $-1$ you need to do a cut different from the standard one that removes the negative $x$-semiaxis.