$\left(x-\displaystyle\frac{x^3}{6}+\displaystyle\frac{x^5}{120} +o(x^5)\right)\left(1+\displaystyle\frac{x^2}{2}-\displaystyle\frac{x^4}{24}+ \displaystyle\frac{x^4}{8}+o(x^4)\right) = x+\displaystyle\frac{x^3}{3}+\displaystyle\frac{2x^5}{15}+o(x^5)$
I got $\ x+ \displaystyle\frac{x^3}{3}+\displaystyle\frac{x^5}{120}-\displaystyle\frac{7x^7}{720}+\displaystyle\frac{x^9}{1440} $
How am I going to get $\displaystyle\frac{2}{15}x^5$? I know that $x^7=o(x^5),x^9=o(x^5)$, but how do coefficients behave?
Coefficients behave as normal (up to and including $x^5$ - the $o(x^5)$ term can conceal any amount of higher-order terms), so either the question is incorrect or you copied down the question wrong. In particular you have two consecutive terms with $x^4$ in the second part - are you sure those are both correct?