Answer
$\frac{1}{16x^{2}}$
Work Step by Step
According to the negative exponent rule, $a^{-n}=\frac{1}{a^{n}}$.
Therefore, $(4x)^{-2}=\frac{1}{(4x)^{2}}=\frac{1}{4^{2}x^{2}}=\frac{1}{16x^{2}}$.
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