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(m^3*n)^-3/m^-7*n^8
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\[\frac{(m^3n)^{-3}}{m^{-7}n^8}\] like that?
\[\frac{(m^3n)^{-3}}{m^{-7}n^8}\] first multiply the exponents in the numerator to get rid of the parentheses \[\frac{m^{-9}n^{-3}}{m^{-7}n^8}\]
then since no doubt you want positive exponents only, you are going to bring everything to to denominator and change the signs of the exponents \[\frac{1}{m^9n^3m^{-7}n^8}\]
then you are going to add the exponents \[\frac{1}{m^{2}n^{11}}\] is your final answer
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