Answer
$
f(x)=3\left(\frac{1}{9}\right)^x
$
Work Step by Step
We know that $f(2)=a b^{2}=\frac{1}{27}$ and $f(-1)=a b^{-1}=27$. The ratio method gives:
$$
\begin{aligned}
\frac{a b^{2}}{a b^{-1}} & =\frac{f(2)}{f(-1)} \\
b^{3} & =\frac{1/27}{27} \\
b & =\left(\frac{1}{27^2}\right)^{1 / 3} =\frac{1}{9}
\end{aligned}
$$ We determine $a$: $$
\begin{aligned}
a\left(\frac{1}{9}\right)^{-1} & =27 \\
9a & =27 \\
a & =3
\end{aligned}
$$ It follows that $$
f(x)=3\left(\frac{1}{9}\right)^x
$$
