Consider $f(x)$ to be a function with it's Taylor series given by $f(x) = 1-x+3x^2 + o(x^2)$ as $x \to 0$. To write down the second order Taylor polynomial of $\ln(f)$ I met some difficulties. Indeed
$$\ln(f) = \ln(1-x + 3x^2 + o(x^2))$$
and to work with Taylor Polynomial I need to understand how $o(x)$ works when inside a function or when derived.
$$\ln(f)\big|_{x = 0} = \ln(1 + o(0)) = \ln(1) = 0$$
What is exactly $o(0)$? I treated it as zero...
Now
$$\text{first derivative} = \frac{-1+6x + o'(x^2)}{1-x+3x^2 + o(x^2)}$$
What is $o'(x^2)$? How to treat it and the other little $-o$ in the denominator? Thank you !!