曲线 $y = \text{f}(x)$ 满足 $\text{f}''(x) = 12\text{e}^{3x} + 7$，且点 $A\left(0, -\frac{2}{3}\right)$ 在该曲线上。 曲线在点 $A$ 处的斜率为 9。求该曲线的方程。

1. Integrate the second derivative to find the first derivative, f'(x).

$\text{f}'(x) = \int (12\text{e}^{3x} + 7) \, \text{d}x$
$\text{f}'(x) = \frac{12\text{e}^{3x}}{3} + 7x + c$
$\text{f}'(x) = 4\text{e}^{3x} + 7x + c$

2. Use the given gradient at x = 0 to find the constant of integration c.

Given gradient at $A(0, -\frac{2}{3})$ is 9, so $\text{f}'(0) = 9$.
$4\text{e}^{3(0)} + 7(0) + c = 9$
$4(1) + c = 9$
$c = 5$
So, $\text{f}'(x) = 4\text{e}^{3x} + 7x + 5$

3. Integrate the first derivative to find the function f(x).

$\text{f}(x) = \int (4\text{e}^{3x} + 7x + 5) \, \text{d}x$
$\text{f}(x) = \frac{4\text{e}^{3x}}{3} + \frac{7x^2}{2} + 5x + d$

4. Use the coordinates of point A to find the constant of integration d.

Since $A(0, -\frac{2}{3})$ lies on the curve, $\text{f}(0) = -\frac{2}{3}$.
$\frac{4}{3}\text{e}^{3(0)} + \frac{7(0)^2}{2} + 5(0) + d = -\frac{2}{3}$
$\frac{4}{3}(1) + d = -\frac{2}{3}$
$d = -\frac{2}{3} - \frac{4}{3} = -\frac{6}{3} = -2$
So, the equation of the curve is $y = \frac{4}{3}\text{e}^{3x} + \frac{7}{2}x^2 + 5x - 2$