To find the \(x\)-coordinate where the \(x\)- and \(y\)-coordinates are increasing at the same rate, we need \(\frac{dy}{dx} = 1\).

First, differentiate \(y = \frac{1}{60}(3x + 1)^2\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{1}{60} \times 2(3x + 1) \times 3 = \frac{1}{10}(3x + 1)\).

Set \(\frac{dy}{dx} = 1\):

\(\frac{1}{10}(3x + 1) = 1\).

Solving for \(x\):

\(3x + 1 = 10\)

\(3x = 9\)

\(x = 3\)