To find \(f(x)\), we need to integrate \(f'(x) = \frac{1}{\sqrt{x+6}} + \frac{6}{x^2}\).

The integral of \(\frac{1}{\sqrt{x+6}}\) is \(2\sqrt{x+6}\).

The integral of \(\frac{6}{x^2}\) is \(-\frac{6}{x}\).

Thus, \(f(x) = 2\sqrt{x+6} - \frac{6}{x} + c\).

Using the condition \(f(3) = 1\), substitute \(x = 3\) into the equation:

\(2\sqrt{3+6} - \frac{6}{3} + c = 1\).

This simplifies to \(2(3) - 2 + c = 1\).

\(6 - 2 + c = 1\).

\(c = -3\).

Therefore, \(f(x) = 2\sqrt{x+6} - \frac{6}{x} - 3\).