Answer
$-x+\dfrac{7x^3}{6}+...$
Work Step by Step
We have the Maclaurin series for $ \sin x$ as follows:
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$
Replace $(x^3-x)$ for $x$ in the above form.
Therefore, we have:
$f(x)=\sin (x^3-x)=(x^3-x)-\dfrac{(x^3-x)^3}{3!}+...\\=(x^3-x)-\dfrac{x^9-3x^7+3x^5-x^3}{6}+...\\=(x^3-x)-\dfrac{-x^3}{6}+...\\=-x+\dfrac{7x^3}{6}+...$
