Since, if $x > 0$ then $Dx^{r} = rx^{r-1}$ for real $r$, when do we have this result for $x \leq 0$?
I think the point is to circumvent the trouble that if $x \leq 0$ then $\log x$ is meaningless, as $$Dx^{r} = D e^{r\log x} = e^{r\log x}\cdot \frac{r}{x} = rx^{r-1}.$$
It very likely depends on precisely how you are defining $x^r$ on negative $x$ for non-integer $r$. However, we can sort of work around this while maintaining your original idea. In particular, we want to define: $$x^r=e^{\log(x)\cdot r}$$ and the only troublesome bit in here is $\log(x)$, which isn't clearly defined for negative numbers. Let us work around this by defining: $$x^r = e^{f(x)\cdot r}$$ where $f(x)$ is any continuous function such that $e^{f(x)}=x$. Differentiating with $x$, one determines that: $$e^{f(x)}f'(x)=1$$ $$f'(x)=e^{-f(x)}$$ Then, if we differentiate $x^r=e^{f(x)\cdot r}$ we get: $$e^{f(x)\cdot r}\cdot f'(x)\cdot r$$ and replacing $f'(x)$ by the identity we found for it: $$e^{f(x)\cdot r}\cdot e^{-f(x)}\cdot r$$ or simplifying: $$r\cdot e^{f(x)\cdot (r-1)}=r\cdot x^{r-1}$$ as desired.
Okay, this might seem a bit suspicious at first - $f$ is really just what we want to call a logarithm - and somehow, we're claiming this works for negative numbers too, even though it's basically the same as the work you did. The subtlety is that all of the above works perfectly well in the complex plane (except at $0$, which can easily be handled alone), but is not as straightforwards as you'd hope. In particular, we can take the logarithm of any non-zero complex number, but since $e^x$ is periodic in $2i\pi$ (see Euler's identity), there is no unique inverse - thus $f$ stands in for what the logarithm is doing "near $x$" - so we're saying no matter how we locally choose our inverse, the theorem works out. We could, indeed, define $\log(-x)=\log(x)+i\pi$ (for $x>0$), and do all the above directly with that definition, but the above shows that as long as our definition of $x^r$ is sensible, things will work out okay.