To solve the equation
\[ \frac{3^{x+1} - 3^x}{3^{x-1}} = 2\left(\frac{1}{9}\right)^{x-1} \]
first, we simplify the left-hand side:
\[ \frac{3^{x+1} - 3^x}{3^{x-1}} = \frac{3^x \cdot 3 - 3^x}{3^{x-1}} = \frac{3^x (3 - 1)}{3^{x-1}} = \frac{3^x \cdot 2}{3^{x-1}} = 2 \cdot \frac{3^x}{3^{x-1}} = 2 \cdot 3 = 6. \]
Next, we simplify the right-hand side:
\[ 2\left(\frac{1}{9}\right)^{x-1} = 2\left(\frac{1}{3^2}\right)^{x-1} = 2\cdot\frac{1}{3^{2(x-1)}} = 2\cdot\frac{1}{3^{2x - 2}} = \frac{2}{3^{2x-2}}. \]
Now we rewrite the equation with our simplifications:
\[ 6 = \frac{2}{3^{2x-2}}. \]
Next, we cross-multiply:
\[ 6 \cdot 3^{2x-2} = 2. \]
Dividing both sides by 2 gives:
\[ 3^{2x-2} = \frac{1}{3}. \]
This can be rewritten as:
\[ 3^{2x-2} = 3^{-1}. \]
Since the bases are the same, we equate the exponents:
\[ 2x - 2 = -1. \]
Now, solving for \(x\):
\[ 2x = -1 + 2 \ 2x = 1 \ x = \frac{1}{2}. \]
Thus, the solution is:
\[ \boxed{\frac{1}{2}}. \]