#### Answer

$\dfrac{1}{9}$

#### Work Step by Step

In order to solve the given problem, we will use the following two rules:
$ (a) a^{-p}=\dfrac{1}{a^p} \\ (b) a^{pq}=(a^p)^q$
We will use Rule-(b) as: $4^{-x}=(2^2)^{-x}=2^{-2x}$
Now, we will use Rule-(a) as: $2^{-2x}=\dfrac{1}{2^{2x}}$
So, $4^{-x}=\dfrac{1}{(2^x)^2}=(2^x)^{-2}$
Since, $2^x=3$, then we simplify the expression as: $4^{-x}=3^{-2}$
Therefore, $4^{-x}=\dfrac{1}{3^2} \\ 4^{-x}=\dfrac{1}{9}$