Use the binomial theorem:
[tex]\left(\dfrac13a^2-2b\right)^6=\displaystyle\sum_{k=0}^6\binom6k\left(\dfrac13a^2\right)^{6-k}(-2b)^k=\sum_{k=0}^6\binom6k\dfrac{(-6)^k}{729}a^{12-2k}b^k[/tex]
where
[tex]\dbinom nk=\dfrac{n!}{k!(n-k)!}[/tex]
is the binomial coefficient.
We get the a⁴ b⁴ term when k = 4, which gives the coefficient
[tex]\dbinom64\dfrac{(-6)^4}{729}=\boxed{\dfrac{80}3}[/tex]
