help me pls

It would be good if another mathematician checks this.  I think it is ok. 

Heureka taught me just a little while ago.

Find the coefficient of   \(a^4b^4\)   in the expansion of   \((\frac{a^2}{3}-2b)^6\)

what you need first is the coefficient of      \((\frac{a^2}{3})^2\;(-2b)^4\)

it will be      \(\frac{6!}{2!4!}=15\) 

so we have       \(15\cdot (\frac{a^2}{3})^2\;(-2b)^4\)

\(15\cdot (\frac{a^2}{3})^2\;(-2b)^4\\ =15\cdot (\frac{a^4}{9})\;\cdot 16b^4\\ =5\cdot (\frac{a^4}{3})\;\cdot 16b^4\\ =\frac{80}{3} a^4b^4\\ \)

so the coefficient is      \(\frac{80}{3}\)

This looks good to me!