[tex]\left(\dfrac{3a^{-3}b^2}{2a^{-1}b^0}\right)^2[/tex]
First some simplifying:
[tex]\left(\dfrac{3a^1b^2}{2a^3b^0}\right)^2=\left(\dfrac{3b^2}{2a^2}\right)^2=\dfrac{9b^4}{4a^4}[/tex]
Then with [tex]a=-2[/tex] and [tex]b=-3[/tex], we have [tex]a^4=(-2)^4=16=4^2[/tex] and [tex]b^4=(-3)^4=81=9^2[/tex], so
[tex]\dfrac{9b^4}{4a^4}=\dfrac{9\cdot9^2}{4\cdot4^2}=\dfrac{9^3}{4^3}=\dfrac{729}{64}[/tex]
