Solve the equation: \( \frac{1}{x}+\frac{2}{3 x^{2}}=\frac{1}{2}+\frac{1}{3 x} \)

To solve the equation \[ \frac{1}{x}+\frac{2}{3 x^{2}}=\frac{1}{2}+\frac{1}{3 x}, \] let's start by finding a common denominator for the left side and the right side. The common denominator for the left side is \(3x^2\), and for the right side, it will be \(6x\). **Step 1: Rewrite both sides.** The left side becomes: \[ \frac{3}{3x} + \frac{2}{3x^2} = \frac{3}{3x} + \frac{2}{3x^2} = \frac{3x + 2}{3x^2}. \] The right side becomes: \[ \frac{3x}{6x} + \frac{2}{6x} = \frac{3x + 2}{6x}. \] Now we have: \[ \frac{3x + 2}{3x^2} = \frac{3x + 2}{6x}. \] **Step 2: Cross multiplying.** To eliminate the fractions, we can cross-multiply: \[ (3x + 2) \cdot 6x = (3x + 2) \cdot 3x^2. \] **Step 3: Condition for valid operations.** We must remember that we cannot divide by zero, so we set a condition: \(3x + 2 \neq 0\) which implies \(x \neq -\frac{2}{3}\). **Step 4: Simplify the equation.** Assuming \(3x + 2 \neq 0\), we can cancel \(3x + 2\) from both sides: \[ 6x = 3x^2. \] **Step 5: Rearranging terms.** This leads us to: \[ 3x^2 - 6x = 0. \] Factoring out \(3x\) gives us: \[ 3x(x - 2) = 0. \] **Step 6: Finding solutions.** Setting each factor equal to zero gives us: \[ 3x = 0 \quad \Rightarrow \quad x = 0, \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2. \] However, we recall the condition that \(x \neq 0\) (division by \(x\) in the original equation). Hence, the only valid solution is: \[ \boxed{2}. \]