Logs on both sides!

$\begin{aligned} \log_3\left(\log_9 x \right) & = \log_9\left(\log_3 x \right) & \small \color{#3D99F6} \text{Using }\log_3 \text{ throughout} \\ \log_3\left(\frac {\log_3 x}{\color{#3D99F6}\log_3 9} \right) & = \frac {\log_3\left(\log_3 x \right)}{\color{#3D99F6} \log_3 9} & \small \color{#3D99F6} \text{Note that }\log_3 9 = \log_3 3^2 = 2 \log_3 3 = 2 \\ \log_3\left(\frac {\log_3 x}{\color{#3D99F6}2} \right) & = \frac {\log_3\left(\log_3 x \right)}{\color{#3D99F6} 2} \\ \log_3\left(\frac {\log_3 x}2 \right) & = \log_3\left(\log_3 x \right)^\frac 12 \\ \implies \frac {\log_3 x}2 & = \sqrt{\log_3 x} & \small \color{#3D99F6} \text{Squaring both sides and rearranging} \\ \log_3^2 x & = 4 \log_3 x & \small \color{#3D99F6} \text{Since }\log_3 x \ne 0 \\ \log_3 x & = 4 \\ \implies x & = 3^4 = \boxed {81} \end{aligned}$

$log3(log9(x))=log9(log3(x))$

let,

$log9(x)=y [1]$

$*9^y=x;$

$log9(log3(x))=z$

$log3(x)=log3(9^y)=log3((3^2)^y)=2y$

$log9(2y)=z$

[2] $log9(2y)$ = $log3(y)$ and let $log3(y)=b$

then [3] $3^b=y$

and $log9(2y)=b$ , refer[2]

$9^b=2y$

refer[3] $(2y=2*3^b, and 2y=9^b)$

$2*3^b=9^b ,9^b=(3^2)^b=(3^b)^2$

y=3^b.

2*y=y^2

$y^2=2y$

y=2;

$log9(x)=2$

$x=9^2=81.$

$\log_3 (\log_9 x) = \log_9 (\log_3 x)$

$\log_3 (\log_9 x) = \frac{\log_3 (\log_3 x)}{\log_3 9}$

$\log_3 (\log_9 x) = \frac{\log_3 (\log_3 x)}{2}$

$2 \log_3 (\log_9 x) = \log_3 (\log_3 x)$

$\log_3 (\log_9 x)^2 = \log_3 (\log_3 x)$

$(\log_9 x)^2 = \log_3 x$

$(\frac{\log_3 x}{\log_3 9})^2 = \log_3 x$

$(\frac{\log_3 x}{2})^2 = \log_3 x$

$\frac{(\log_3 x)^2}{4} = \log_3 x$

$\frac{(\log_3 x)^2}{\log_3 x} = 4$

$\log_3 x = 4$

$x = 3^4$

$x = \boxed{81}$