A curve is such that $\frac{d^2y}{dx^2} = 6e^{3x} - x$ and passes through the point $\left(2, \frac{2}{3}e^6\right)$.

1. Integrate the second derivative to find the expression for $\frac{dy}{dx}$.

$\frac{dy}{dx} = \int (6e^{3x} - x) \, dx$
$\frac{dy}{dx} = 2e^{3x} - \frac{1}{2}x^2 + c$

2. Use the given condition to find the constant of integration $c$.

Given $\frac{dy}{dx} = 5$ when $x = 0$:
$5 = 2e^{3(0)} - \frac{1}{2}(0)^2 + c$
$5 = 2(1) + c$
$c = 3$
So, $\frac{dy}{dx} = 2e^{3x} - \frac{1}{2}x^2 + 3$

3. Integrate $\frac{dy}{dx}$ to find the expression for $y$.

$y = \int \left(2e^{3x} - \frac{1}{2}x^2 + 3\right) \, dx$
$y = \frac{2}{3}e^{3x} - \frac{1}{6}x^3 + 3x + d$

4. Use the given point to find the constant of integration $d$ and state the final equation.

Given the curve passes through $\left(2, \frac{2}{3}e^6\right)$:
$\frac{2}{3}e^6 = \frac{2}{3}e^{3(2)} - \frac{1}{6}(2)^3 + 3(2) + d$
$\frac{2}{3}e^6 = \frac{2}{3}e^6 - \frac{8}{6} + 6 + d$
$0 = -\frac{4}{3} + 6 + d$
$d = \frac{4}{3} - 6 = -\frac{14}{3}$

Equation of the curve: $y = \frac{2}{3}e^{3x} - \frac{1}{6}x^3 + 3x - \frac{14}{3}$

1. Find the expression for $\frac{dy}{dx}$ by integrating $\frac{d^2y}{dx^2}$ and using the given condition

$\frac{dy}{dx} = \int (6e^{3x} - x) \, dx = 2e^{3x} - \frac{1}{2}x^2 + C$
Given $\frac{dy}{dx} = 5$ when $x = 0$:
$5 = 2e^0 - 0 + C \implies C = 3$
$\frac{dy}{dx} = 2e^{3x} - \frac{1}{2}x^2 + 3$

2. Calculate the gradient of the tangent at $x = 2$

When $x = 2$,
$m = \frac{dy}{dx} = 2e^{3(2)} - \frac{1}{2}(2)^2 + 3$
$m = 2e^6 - 2 + 3 = 2e^6 + 1$

3. Find the equation of the tangent line

Using $y - y_1 = m(x - x_1)$ with point $(2, \frac{2}{3}e^6)$:
$y - \frac{2}{3}e^6 = (2e^6 + 1)(x - 2)$
$y = (2e^6 + 1)x - 2(2e^6 + 1) + \frac{2}{3}e^6$
$y = (2e^6 + 1)x - 4e^6 - 2 + \frac{2}{3}e^6$
$y = (2e^6 + 1)x - \frac{10}{3}e^6 - 2$