First, find \(\frac{dy}{dx}\) for the curve \(y = (7x^2 + 1)^{\frac{1}{3}}\).

Using the chain rule, \(\frac{dy}{dx} = \left[ \frac{1}{3} (7x^2 + 1)^{-\frac{2}{3}} \right] \times [14x]\).

When \(x = 3\), \(\frac{dy}{dx} = \frac{1}{3} \times (64)^{-\frac{2}{3}} \times 42\).

Calculate \((64)^{-\frac{2}{3}} = \left( \frac{1}{64} \right)^{\frac{2}{3}} = \left( \frac{1}{4} \right)^2 = \frac{1}{16}\).

Thus, \(\frac{dy}{dx} = \frac{1}{3} \times \frac{1}{16} \times 42 = \frac{7}{8}\).

Given \(\frac{dx}{dt} = 8\), use the chain rule to find \(\frac{dy}{dt}\):

\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = \frac{7}{8} \times 8 = 7\).