Taylor series expansion of $f(x) = \sin^3 \left(\ln(1+x) \right)$. How does one use Taylor series expansion to compute $f^{(3)}(0)$ in which $f(x) = \sin^3 \left(\ln(1+x) \right)$.

Probably easier to use a linearized expression for $\sin^3(\theta)$ :
$$\sin^3(\theta)=\frac{3}{4}\sin(\theta)-\frac{1}{4}\sin(3\theta)$$
Now, for all $x>-1$ :
$$f(x)=\sin^3(\ln(1+x))=\frac{3}{4}\sin(\ln(1+x))-\frac{1}{4}\sin(3\ln(1+x))$$
and the Taylor expansion at $x=0$ is :
\begin{align}
f(x)&=&\frac{3}{4}\sin(x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3))-\frac{1}{4}\sin(3x-\frac{3x^2}{2}+x^3+o(x^3))\\
&=&\frac{3}{4}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^3}{6}+o(x^3)\right)-\frac{1}{4}\left(3x-\frac{3x^2}{2}+x^3-\frac{9x^3}{2}+o(x^3)\right)\\
&=&x^3+o(x^3)
\end{align}
Finally : $f^{(3)}(0)=3!\,\times\,1$, that is :$$\boxed{f^{(3)}(0)=6}$$
