To find \(f(x)\), integrate \(f'(x) = (3x - 1)^{-\frac{1}{3}}\).

\(f(x) = \int (3x - 1)^{-\frac{1}{3}} \, dx\)

Using substitution, let \(u = 3x - 1\), then \(du = 3 \, dx\), so \(dx = \frac{du}{3}\).

\(f(x) = \int u^{-\frac{1}{3}} \cdot \frac{1}{3} \, du\)

\(f(x) = \frac{1}{3} \cdot \frac{u^{\frac{2}{3}}}{\frac{2}{3}} + c\)

\(f(x) = \frac{1}{2} (3x - 1)^{\frac{2}{3}} + c\)

Given \(f(3) = 1\), substitute to find \(c\):

\(1 = \frac{1}{2} (3 \cdot 3 - 1)^{\frac{2}{3}} + c\)

\(1 = \frac{1}{2} \cdot 8^{\frac{2}{3}} + c\)

\(1 = \frac{1}{2} \cdot 4 + c\)

\(1 = 2 + c\)

\(c = -1\)

Thus, \(f(x) = \frac{1}{2} (3x - 1)^{\frac{2}{3}} - 1\).

To find the y-coordinate of \(B\), set \(x = 0\):

\(y = \frac{1}{2} (3 \cdot 0 - 1)^{\frac{2}{3}} - 1\)

\(y = \frac{1}{2} (-1)^{\frac{2}{3}} - 1\)

Since \((-1)^{\frac{2}{3}} = \left((-1)^{\frac{1}{3}}\right)^2 = (-1)^2 = 1\),

\(y = \frac{1}{2} \cdot 1 - 1\)

\(y = -\frac{1}{2}\)

Therefore, the y-coordinate of \(B\) is \(-\frac{1}{2}\).