To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = \frac{6}{(3x - 2)^3}\).

Let \(u = 3x - 2\), then \(\frac{du}{dx} = 3\) or \(dx = \frac{du}{3}\).

Substitute into the integral:

\(\int \frac{6}{u^3} \cdot \frac{du}{3} = 2 \int u^{-3} \, du\).

The integral of \(u^{-3}\) is \(\frac{u^{-2}}{-2}\), so:

\(2 \left( \frac{u^{-2}}{-2} \right) = -u^{-2} + c\).

Substitute back \(u = 3x - 2\):

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

Using the point \(A(1, -3)\), substitute \(x = 1\) and \(y = -3\):

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

\(-3 = -1 + c\).

\(c = -2\).

Thus, the equation of the curve is:

\(y = -(3x-2)^{-2} - 2\).