what is the taylor expansion of $\ln(\sin x/x)$ what is the Taylor expansion of the function
$$f(x) = \ln \frac{\sin x}x$$
around  the point $x=0$? Ignore powers of $x$ which are greater than $6$.
Here is my method:
$$\ln(1+x)=x-\frac{x^2}2 + \frac{x^3}3 -\frac{x^4}4,$$
so we should get the function
$$g(x)=\frac{\sin x}x$$
in the form of $1 + x$ to get the answer … or is there another way to consider the Taylor expansion of $\ln (u)$ by considering $f$ to be the answer of $$\frac{\sin x}x\cdots$$
Which way is the best?? What is the final answer???  

As you suggest, it will be useful to know the expansion
$$\ln(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots\tag{1}$$
(at least for $|t|\lt 1$).
We also need the power series expansion for $\sin x$, which is
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.$$
So the function which is $1$ at $x=0$ and $\frac{\sin x}{x}$ elsewhere has power series expansion 
$$1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots.\tag{2}$$
Now substitute (2) for $1+t$ in (1). Thus $t=-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots$.
Make  sure not to compute too far. So throw away all terms in powers of $x$ greater than $6$.
Because the expression (2) has no $x$ term, you will only need to compute to the $\frac{t^3}{3}$ term in (1). So the calculation is less time-consuming than it seems. 
Getting a general term for the power series expansion looks, however, very difficult. Luckily you were not asked to find many terms. You will end up with something of the shape $ax^2+bx^4+cx^6$.
