(i) To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{3}{(1 + 2x)^2}\).

Let \(u = 1 + 2x\), then \(du = 2dx\) or \(dx = \frac{du}{2}\).

\(\int \frac{3}{u^2} \cdot \frac{du}{2} = -\frac{3}{2u} + c\).

Substitute back \(u = 1 + 2x\):

\(y = -\frac{3}{2(1 + 2x)} + c\).

Using the point \((1, \frac{1}{2})\):

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

\(\frac{1}{2} = -\frac{3}{6} + c\).

\(\frac{1}{2} = -\frac{1}{2} + c\).

\(c = 1\).

Thus, the equation of the curve is \(y = \frac{3}{1 + 2x} + 1\).

(ii) We need \(\frac{3}{(1 + 2x)^2} < \frac{1}{3}\).

\(9 < (1 + 2x)^2\).

\((1 + 2x)^2 > 9\).

\(1 + 2x > 3\) or \(1 + 2x < -3\).

\(2x > 2\) or \(2x < -4\).

\(x > 1\) or \(x < -2\).

Thus, the set of values of \(x\) is \(x > 1, x < -2\).