How do you find the Taylor Series of $f(x) = x^{2}*\sin(\frac{x^{2}}{3})$ about $a = 0$? Please without getting too technical, this is calculus 2. I know that the general form of a Taylor Series about 0 (Maclaurin) is: $$\sum_{n=0}^{\infty} \frac{f^{n}(0)}{n!} * x^{n}$$ and so, $$f(x) = f(0)/0! + f (0)/1!*x + f (0)/2!*x^2 + f (0)/3!*x^3 + .... $$
Just very unsure how this one would be carried out and simplified. Is it simply taking derivatives? Can it be put into general summation notation? 
Please help
Thank you

Use Taylor Series for $\sin x$ 
$$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!} -\frac{x^7}{7!}\ldots$$
$$\sin {\left(\frac{x^2}{3} \right)}=x-\frac{{\left(\frac{x^2}{3} \right)}^3}{3!}+\frac{{\left(\frac{x^2}{3} \right)}^5}{5!} -\frac{{\left(\frac{x^2}{3} \right)}^7}{7!}\ldots$$
You want $x^2 \sin {\left(\frac{x^2}{3} \right)}$, so multiply the above series by $x^2$.
$$x^2 \sin {\left(\frac{x^2}{3} \right)} = x^3- \frac{x^8}{162} +\frac{x^{12}} {29160} \ldots$$
In general :
$$x^2 \sin {\left(\frac{x^2}{3} \right)} = \sum _{n=0}^{\infty}\frac{(-1)^n x^{4n+6}}{3^{2n+1} (2n+1)!}$$
