To solve the equation \(( \frac{2}{3}x - \frac{4}{9} ) \cdot ( \frac{1}{2} + \frac{-3}{7} \div x ) = 0\), we need to break it down step by step.

First, let's rewrite the equation for clarity:
\[ \left( \frac{2}{3}x - \frac{4}{9} \right) \cdot \left( \frac{1}{2} + \frac{-3}{7} \div x \right) = 0 \]

### Step 1: Simplify the second term inside the parentheses
The term \(\frac{-3}{7} \div x\) can be rewritten as \(\frac{-3}{7} \cdot \frac{1}{x}\):
\[ \frac{-3}{7} \div x = \frac{-3}{7x} \]

So the equation becomes:
\[ \left( \frac{2}{3}x - \frac{4}{9} \right) \cdot \left( \frac{1}{2} + \frac{-3}{7x} \right) = 0 \]

### Step 2: Combine the terms inside the second parentheses
\[ \frac{1}{2} + \frac{-3}{7x} = \frac{1}{2} - \frac{3}{7x} \]

### Step 3: Set up the equation for zero product property
For the product of two terms to be zero, at least one of the terms must be zero. Therefore, we have two cases to consider:

1. \(\frac{2}{3}x - \frac{4}{9} = 0\)
2. \(\frac{1}{2} - \frac{3}{7x} = 0\)

### Case 1: Solve \(\frac{2}{3}x - \frac{4}{9} = 0\)
\[ \frac{2}{3}x - \frac{4}{9} = 0 \]
Add \(\frac{4}{9}\) to both sides:
\[ \frac{2}{3}x = \frac{4}{9} \]
Multiply both sides by \(\frac{3}{2}\) to solve for \(x\):
\[ x = \frac{4}{9} \cdot \frac{3}{2} = \frac{4 \cdot 3}{9 \cdot 2} = \frac{12}{18} = \frac{2}{3} \]

### Case 2: Solve \(\frac{1}{2} - \frac{3}{7x} = 0\)
\[ \frac{1}{2} - \frac{3}{7x} = 0 \]
Add \(\frac{3}{7x}\) to both sides:
\[ \frac{1}{2} = \frac{3}{7x} \]
Cross-multiply to solve for \(x\):
\[ 1 \cdot 7x = 2 \cdot 3 \]
\[ 7x = 6 \]
Divide both sides by 7:
\[ x = \frac{6}{7} \]

### Conclusion
The solutions to the equation are:
\[ x = \frac{2}{3} \quad \text{or} \quad x = \frac{6}{7} \]