Find the Taylor series expansion (centered at $z_{0}=0$) of the function $f(z)=\sin(z^3)$. Find the Taylor series expansion (centered at $z_{0}=0$) of the function $f(z)=\sin(z^3)$. 
Use this expansion: a) to find $f^{69}(0)$; b) to compute the integral transversed once in the positive (with respect to the disk) direction 
$\oint\limits_{|z|=3}{\frac{f(z)}{z^{70}}}dz$

I know that if a function $f$ is analytic in a disk centered at $z_0$ of radius $r$, then it can be represented as a sum of it's tailor series. The disk here is centered at the origin and has a radius of 3. 
Should I be using the formula to find the Taylor Coefficients at $z_0$: $a_n=\frac{f^{(n)(z_0)}}{n!}$? 
I am looking for help with coming up with the solution to this problem. I feel that I have the correct pieces and am not sure of how to connect them. 

Use the Taylor series for sine (just plug in $z^3$ wherever a $z$ appears in the original series). Notice what this "does" to the coefficients: the coefficient of $z$ in the original series is now the coefficient of $z^3$, the coefficient of $z^2$ in the original now belongs to $z^6$, etc. 
