1. The problem is to factorize the expression $$y^2 - \frac{1}{9}$$. 2. This is a difference of squares, which follows the formula $$a^2 - b^2 = (a - b)(a + b)$$. 3. Here, $$a = y$$ and $$b = \frac{1}{3}$$ because $$\left(\frac{1}{3}\right)^2 = \frac{1}{9}$$. 4. Applying the formula, we get: $$y^2 - \frac{1}{9} = \left(y - \frac{1}{3}\right)\left(y + \frac{1}{3}\right)$$. 5. Therefore, the correct factorization is option B: $$(y - \frac{1}{3})(y + \frac{1}{3})$$. This factorization shows the product of two binomials whose squares give the original expression.