Answer
$
f(x)=2\left(\frac{1}{3}\right)^x
$
Work Step by Step
We know that $f(-3)=a b^{-3}=54$ and $f(2)=a b^2=\frac{2}{9}$. The ratio method gives:
$$
\begin{aligned}
\frac{a b^{2}}{a b^{-3}} & =\frac{f(2)}{f(-3)} \\
b^{5} & =\frac{2/9}{54} \\
b & =\left(\frac{1}{243}\right)^{1 / 5} =\frac{1}{3}
\end{aligned}
$$ We determine $a$: $$
\begin{aligned}
a\left(\frac{1}{3}\right)^2 & =\frac{2}{9} \\
\frac{a}{9} & =\frac{2}{9} \\
a & =2
\end{aligned}
$$ It follows that $$
f(x)=2\left(\frac{1}{3}\right)^x
$$
