Closed form for $(a_n)$ such that $a_{n+2} = \frac{a_{n+1}a_n}{6a_n - 9a_{n+1}}$ with $a_1=1$, $a_2=9$

Hint: Let $b_n=\frac1{a_n}$. Then $b_1=1$, $b_2=\frac19$, and $b_{n+2}=6b_{n+1}-9b_n$ for every $n\geqslant1$. This affine recursion of order $2$ has a double root at $__$ hence $b_n=$ $(An+B)\cdot$ $__$ ${}^n$ for every $n$. Identifying $A$ and $B$ yields finally $a_n=$ $______$ for every $n\geqslant1$.