Bashing Exponential

Put $3^x=t$ and $9^x=t^2$ so that equation simplifies to:

$t^2-4t+6-\dfrac{4}t+\dfrac{1}{t^2}=0$ $\iff \left(t^2+\dfrac{1}{t^2}\right)-4\left(t+\dfrac{1}{t}\right)+6=0$ $\iff\left(t+\frac{1}{t}\right)^2-4\left(t+\frac{1}t\right)+4=0$ $\iff \left(t+\dfrac 1t-2\right)^2=0$ $t+\dfrac 1t=2\iff t=3^x=1\iff \boxed{\color{#D61F06}x=0}$ (Since $t+\dfrac 1t \ge 2~\forall t\ge 0$ )

$\begin{aligned} 9^x - 4 \cdot 3^x + 6 - \frac{4}{3^x} + \frac{1}{9^x} &= 0\\ 9^{-x} \left(3^x -1\right)^4 &=0\\ \end{aligned}$

$9^{-x} = 0$ is impossible. Since we're searching real value of $x$ , there's no $x$ that satisfy the equation.

So the only possible solution is $3^x - 1 = 0$ . Hence $x = \color{#D61F06}{\boxed{0}}$ .