It is given that $f'(x) = \sin 3x - \frac{1}{2x + 1}$ and $f(0) = \frac{2}{3}$. Find an expression for $6f(x) + f''(x)$.

1. Integrate $f'(x)$ to find the general expression for $f(x)$.

$f(x) = \int \left( \sin 3x - \frac{1}{2x + 1} \right) dx$
$f(x) = -\frac{1}{3}\cos 3x - \frac{1}{2}\ln(2x + 1) + C$

2. Use the given condition $f(0) = \frac{2}{3}$ to find the constant of integration $C$.

$f(0) = -\frac{1}{3}\cos(0) - \frac{1}{2}\ln(1) + C = \frac{2}{3}$
$-\frac{1}{3}(1) - 0 + C = \frac{2}{3}$
$C = \frac{2}{3} + \frac{1}{3} = 1$
$f(x) = -\frac{1}{3}\cos 3x - \frac{1}{2}\ln(2x + 1) + 1$

3. Differentiate $f'(x)$ to find $f''(x)$.

$f'(x) = \sin 3x - (2x + 1)^{-1}$
$f''(x) = \frac{d}{dx}(\sin 3x) - \frac{d}{dx}((2x + 1)^{-1})$
$f''(x) = 3\cos 3x - (-1)(2x + 1)^{-2}(2)$
$f''(x) = 3\cos 3x + \frac{2}{(2x + 1)^2}$

4. Substitute $f(x)$ and $f''(x)$ into the expression $6f(x) + f''(x)$ and simplify.

$6f(x) + f''(x) = 6\left( -\frac{1}{3}\cos 3x - \frac{1}{2}\ln(2x + 1) + 1 \right) + \left( 3\cos 3x + \frac{2}{(2x + 1)^2} \right)$
$= -2\cos 3x - 3\ln(2x + 1) + 6 + 3\cos 3x + \frac{2}{(2x + 1)^2}$
$= \cos 3x - 3\ln(2x + 1) + \frac{2}{(2x + 1)^2} + 6$