\((3x^{-3})^2(-2xy)}\)
The square (²) can be distributed to each factor inside those parentheses.
3² = 3 × 3 = 9
as for the x^-3, we want to multiply our exponents together, leaving x^-6.
\((9x^{-6})(-2xy)\)
Note that since multiplication is commutative/associative, we can multiply these together in whatever order to simplify.
The x on the right can be rewritten as x¹. Multiply that and x^-6 together to get x^-5. (we just add the exponents)
As for the 9 and -2, multiply those together to get -18.
Now we have \(-18x^{-5}y\).
Lastly, if you don't want a negative exponent in your answer:
It can be flipped to the bottom because of this property:
\(x^{-n}=\frac{1}{x^n}\)
This leaves us with \(\boxed{\frac{-18y}{x^5}}\)
