Find the $f^{(3)}(0)$ of $f(x) = \sin^{3}(\ln(1+x))$ using Taylor expansion For the Taylor expansions I used:
$\ln(1+x) = \sum_{k=1}^\infty (-1)^{k+1} \frac{x^{k}}{k}$
$\sin(x) = \sum_{k=0}^\infty (-1)^{k+1} \frac{x^{2k+1}}{(2k+1)!}$
So $\sin(x) = x - \frac{x^{3}}{3!} + O(x^{5})$
and $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$ 
Then the composite expansion (for rest of question dropping H.O.T.):
$\sin(\ln(1+x)) \approx (x - \frac{x^2}{2} + \frac{x^3}{3}) - \frac{(x - \frac{x^2}{2} + \frac{x^3}{3})^3}{3!} $
Reducing:
$\approx (x - \frac{x^2}{2} + \frac{x^3}{3}) - \frac{x^3}{6} \approx x - \frac{x^2}{2} + \frac{x^3}{6}$
Cube the result and simplify again:
$(x - \frac{x^2}{2} + \frac{x^3}{6})^3 \approx x^3$ keeping the third-order term.
At this point, my understanding is that the coefficient of the third-order term in the formula for the general Taylor series is $f^{(3)}$
$\frac{f^{(3)}(a)}{3!}(x-a)^3$
So I believe I can infer that the coefficient of the composite Taylor expansion, 1, also relates to $f^{(3)}$, and then solving:
$\frac{f^{(3)}}{3!} = 1$
$f^{(3)} = 3! = 6$
Have I understood this question and the right approach for solving it?

This is correct, though take note that you should be more careful with your remainders on each step to ensure you didn't miss anything that might contribute to the $x^3$ coefficient:
\begin{align}\sin(\ln(1+x))&=\sin\bigg(x-\frac{x^2}2+\frac{x^3}3+\mathcal O(x^4)\bigg)\tag{$\star$}\\&=\left[\color{#2255FF}{x-\frac{x^2}2+\frac{x^3}3+\mathcal O(x^4)}\right]-\frac16(\color{#228855}{x+\mathcal O(x^2)})^3+\mathcal O(\color{#FF9944}{x}^4)\\&=x-\frac{x^2}2+\frac16x^3+\mathcal O(x^4)\end{align}
$(\star)$: $\color{#2255FF}{x-\frac{x^2}2+\frac{x^3}3+\mathcal O(x^4)}=\color{#228855}{x+\mathcal O(x^2)}=\color{#FF9944}{\mathcal O(x)}$
