$\Big[\dfrac{x^{-2}}{3y^{-1}}\Big]^{-1}$ is equal to:

1. $\dfrac{3x^2}{y}$

2. $\dfrac{x^2}{3y}$

3. $\dfrac{y}{3x^2}$

4. $\dfrac{3y}{x^2}$

As we know, for any non-zero rational number a

$a^{-n} = \dfrac{1}{a^n}$ and $a^{n} = \dfrac{1}{a^{-n}}$.

$\Big[\dfrac{x^{-2}}{3y^{-1}}\Big]^{-1}\\[1em] = \Big[\dfrac{y^1}{3x^2}\Big]^{-1}\\[1em] = \Big[\dfrac{3x^2}{y}\Big]$

Hence, option 1 is the correct option.