To find "the other root" it often helps to express the original problem as deviation from that root which is already known. So if we "see" one root is $\small x=3$ I'd proceed $\small 3^x = 9x \to 3^{3+d} = 9(3+d) $ where $\small d=0$ refers to the already known solution. Then the equation often can be much simplified to exhibit a possible interval for another solution more visibly.
$\qquad \small \begin{eqnarray} 3^{3+d} &=& 9(3+d) \\ 27 \cdot 3^d &=& 27+9d \\ 3^d &=& 1+d/3 \\ \end{eqnarray} $
and search for solutions in $\small d \ne 0 $ If we want to avoid calculus here, but know the expression for the exponential series, we might introduce the symbol $\small \lambda = \ln(3) $ and write
$\qquad \small \begin{eqnarray} 1+ \lambda d + (\lambda d)^2/2! + \ldots &=& 1 + d/3 \\ \lambda d + (\lambda d)^2/2! + \ldots &=& d/3 \\ \lambda + \lambda^2 d/2! + \lambda^3 d^2/3! + \ldots &=& 1/3 \\ d( \lambda^2 /2! + \lambda^3 d/3! + \ldots ) &=& 1/3 - \lambda & \lt & 0\\ \end{eqnarray} $
so d must be negative. Then we can write using -c=d and c positive:
$\qquad \small \begin{eqnarray} 3^{-c} &=& 1-c/3 \\ 1/3^c + c/3 &=& 1 \\ \end{eqnarray} $
and have an initial guess ($\small 2 \lt c \lt 3 $) for the interval for the second possible c ($\small c \ne 0 $) resp the second root $\small d = -c \ne 0 \to 0 \lt x \lt 1 $ and might apply Newton's iteration for approximation of the actual root. Also we see immediately that there is no further real (non-complex) root.