First, find the derivative \(\frac{dy}{dx}\) of the curve \(y = \frac{4}{(3x + 1)^2}\).

Using the chain rule, let \(u = 3x + 1\), then \(y = 4u^{-2}\).

The derivative \(\frac{dy}{du} = -8u^{-3}\).

Since \(u = 3x + 1\), \(\frac{du}{dx} = 3\).

Thus, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -8u^{-3} \cdot 3 = -24(3x + 1)^{-3}\).

When \(x = -1\), \(y = \frac{4}{(3(-1) + 1)^2} = \frac{4}{1} = 1\).

Also, \(\frac{dy}{dx} = -24(3(-1) + 1)^{-3} = 3\).

The equation of the tangent line at \(x = -1\) is given by the point-slope form: \(y - y_1 = m(x - x_1)\).

Substitute \(y_1 = 1\), \(m = 3\), and \(x_1 = -1\):

\(y - 1 = 3(x + 1)\).

Simplifying gives \(y = 3x + 4\).