In wolfram alpha I got the following result for the Taylor expansion of $\log(x - x^2)$ at 0:
$$\log(x - x^2) = \log(x)-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\frac{x^6}{6}+O(x^7)$$
I don't see where this comes from? Furthermore, $\log$ is singular at $0$ so how is a Taylor expansion even possible? It seems from this plot that having $x - x^2$ instead of just $x$ as the argument makes the function 'less singular', does this have something to do with it?
(I suppose you mean the natural logarithm when you write $\log$.)
Hint: $\log(x-x^2) = \log x + \log(1-x)$. If you're familiar with the Taylor expansion of $\log$ at $1$, then that's all.
Remember that $\log(x-x^2)$ can't have a classical Taylor expansion at $0$ as it is not defined at $0$. You need the initial term to correct for this.