To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = \frac{4}{(x-3)^3}\).

Integrate: \(y = \int \frac{4}{(x-3)^3} \, dx\).

Let \(u = x-3\), then \(du = dx\).

\(y = \int \frac{4}{u^3} \, du = 4 \int u^{-3} \, du\).

Integrating, we get \(y = 4 \left( \frac{u^{-2}}{-2} \right) + c = \frac{-2}{u^2} + c\).

Substitute back \(u = x-3\): \(y = \frac{-2}{(x-3)^2} + c\).

Use the point (4, 5) to find \(c\):

\(5 = \frac{-2}{(4-3)^2} + c\).

\(5 = -2 + c\).

\(c = 7\).

Thus, the equation of the curve is \(y = \frac{-2}{(x-3)^2} + 7\).