It is given that $f(x)$ is such that $f'(x) = \sin 2x + \cos 3x$. Given also that $f\left(\frac{\pi}{6}\right) = 0$, show that $f''(x) + 9f(x) = -\frac{3}{4} - \frac{5}{2}\cos 2x$.

1. Find the second derivative $f''(x)$ by differentiating $f'(x)$.

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

2. Find the general expression for $f(x)$ by integrating $f'(x)$.

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

3. Use the given condition $f\left(\frac{\pi}{6}\right) = 0$ to find the constant of integration $C$.

$f\left(\frac{\pi}{6}\right) = -\frac{1}{2}\cos\left(2 \cdot \frac{\pi}{6}\right) + \frac{1}{3}\sin\left(3 \cdot \frac{\pi}{6}\right) + C = 0$
$-\frac{1}{2}\cos\left(\frac{\pi}{3}\right) + \frac{1}{3}\sin\left(\frac{\pi}{2}\right) + C = 0$
$-\frac{1}{2}\left(\frac{1}{2}\right) + \frac{1}{3}(1) + C = 0$
$-\frac{1}{4} + \frac{1}{3} + C = 0$
$C = \frac{1}{4} - \frac{1}{3} = -\frac{1}{12}$
So, $f(x) = -\frac{1}{2}\cos 2x + \frac{1}{3}\sin 3x - \frac{1}{12}$.

4. Substitute $f''(x)$ and $f(x)$ into the expression $f''(x) + 9f(x)$ and simplify to show the required result.

$f''(x) + 9f(x) = (2\cos 2x - 3\sin 3x) + 9\left(-\frac{1}{2}\cos 2x + \frac{1}{3}\sin 3x - \frac{1}{12}\right)$
$= 2\cos 2x - 3\sin 3x - \frac{9}{2}\cos 2x + 3\sin 3x - \frac{9}{12}$
$= \left(2 - \frac{9}{2}\right)\cos 2x + (-3 + 3)\sin 3x - \frac{3}{4}$
$= -\frac{5}{2}\cos 2x - \frac{3}{4}$
$= -\frac{3}{4} - \frac{5}{2}\cos 2x \quad \text{(Shown)}$