Undeleted and edited. This idea is an expansion of Ross Millikan's comment. If there's any error the fault is mine.
The first few terms are
$\begin{eqnarray*} T(1) &=&T(2)=U(1)=1 \\ T(3) &=&T(4)=U(2)=12 \\ T(5) &=&T(6)=U(3)=45 \\ &&\cdots \end{eqnarray*}$
Generalizing, we have
$T(2n-1)=T(2n)=U(n),\qquad n\ge 1.\qquad (1)$
Since
$T(2n-1)=3T(2n-3)+9=3T(2n-2)+9,$
we obtain
$U(1)=1,\qquad U(n)=3U(n-1)+9,\qquad n\ge 2.\qquad (2)$
We will apply the rudiments of the theory of recurrences (or difference equations, see e.g. An introduction to difference equations by Saber Elaydi). This is an nonhomogeneous recurrence relation, consequently we have to find a particular solution of the recurrence and solve the homogeneous recurrence, which is linear and has constant coefficients. Hence, the explicit solution is of the form
$U(1)=1,\qquad U(n)=AX^{n}+C,\qquad (3)$
where $X$ is the root of the characteristic equation $X-3=0$ associated to the homogeneous equation $U(n)-3U(n-1)=0$ and $C$ is a particular solution (a constant, in this case) of $U(n)=3U(n-1)+9$. From $(2)$, we must have $C=3C+9$, i.e. $C=-9/2$. So $U(n)=3^{n}A-9/2$.
From the initial conditions $U(1)=1$, we get $1=3A-9/2$, and $A=11/6$. Thus the solution of $(2)$ is
$U(n)=\frac{11}{6}3^{n}-\frac{9}{2},\qquad n\ge 1.\qquad (4)$
Thus the solution of the given recurrence $T(n)=3T(n-2)+9$ is
$T(2n-1)=T(2n)=U(n)=\frac{11}{6}3^{n}-\frac{9}{2},\qquad n\ge 1.\qquad (5)$