一曲线的斜率函数为 $3\cos 4x - \sec^2 3x$。已知该曲线穿过点 $\left(\frac{\pi}{12}, \frac{\sqrt{3}}{4} - \frac{1}{3}\right)$，求该曲线的方程。

1. Integrate the gradient function to find the general equation of the curve

$\frac{dy}{dx} = 3\cos 4x - \sec^2 3x$

$y = \int (3\cos 4x - \sec^2 3x) \, dx$
$y = \frac{3}{4}\sin 4x - \frac{1}{3}\tan 3x + C$

2. Substitute the given coordinates to find the constant of integration $C$

Substitute $x = \frac{\pi}{12}$ and $y = \frac{\sqrt{3}}{4} - \frac{1}{3}$:

$\frac{\sqrt{3}}{4} - \frac{1}{3} = \frac{3}{4}\sin\left(4 \times \frac{\pi}{12}\right) - \frac{1}{3}\tan\left(3 \times \frac{\pi}{12}\right) + C$
$\frac{\sqrt{3}}{4} - \frac{1}{3} = \frac{3}{4}\sin\left(\frac{\pi}{3}\right) - \frac{1}{3}\tan\left(\frac{\pi}{4}\right) + C$

3. Evaluate the trigonometric ratios and solve for $C$

$\frac{\sqrt{3}}{4} - \frac{1}{3} = \frac{3}{4}\left(\frac{\sqrt{3}}{2}\right) - \frac{1}{3}(1) + C$
$\frac{\sqrt{3}}{4} - \frac{1}{3} = \frac{3\sqrt{3}}{8} - \frac{1}{3} + C$
$\frac{2\sqrt{3}}{8} = \frac{3\sqrt{3}}{8} + C$
$C = \frac{2\sqrt{3}}{8} - \frac{3\sqrt{3}}{8} = -\frac{\sqrt{3}}{8}$

The equation of the curve is $y = \frac{3}{4}\sin 4x - \frac{1}{3}\tan 3x - \frac{\sqrt{3}}{8}$.