(2xy^{3})^{-2}

To make a negative exponent positive, move the base and its exponent to the opposite location in the fraction (reciprocation). That is, a^{-n} = 1/a^{n} and 1/a^{-n} = a^{n}.

Using the laws of exponents, we can prove the above by the following:

a^{n} • a^{-n} = a^{n+(-n)} = a^{0} = 1 ; therefore, a^{-n} = 1/a^{n} (where a≠0).

With this, (2xy^{3})^{-2} = 1/(2xy^{3})^{2}

1/(2xy^{3})^{2} = 1/(2^{2}x^{2}(y^{3})^{2})

= 1/(4x^{2}y^{6})