To solve the integral

$\iint_{R} \left( x^2 + 5xy + 6y^2 \right) \, dx \, dy$

over the region $R$ bounded by the lines $y = -\frac{1}{2}x + 3$, $y = -\frac{1}{2}x + 4$, $y = -\frac{1}{3}x$, and $y = -\frac{1}{3}x + 3$, we will use the given transformation:

$u = 2x + 4y, \quad v = x + 3y.$

Step 1: Find the Jacobian of the Transformation

The Jacobian determinant $J$ is calculated by differentiating $u$ and $v$ with respect to $x$ and $y$:

$\frac{\partial u}{\partial x} = 2, \quad \frac{\partial u}{\partial y} = 4,$ $\frac{\partial v}{\partial x} = 1, \quad \frac{\partial v}{\partial y} = 3.$

Then, the Jacobian $J$ is:

$J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} = \begin{vmatrix} 2 & 4 \\ 1 & 3 \end{vmatrix} = (2)(3) - (4)(1) = 6 - 4 = 2.$

Thus, $dx \, dy = \frac{1}{2} \, du \, dv$.

Step 2: Express $x$ and $y$ in Terms of $u$ and $v$

We have the system of equations: $u = 2x + 4y,$ $v = x + 3y.$

Solving for $x$ and $y$, we can use substitution or matrix inversion. I'll proceed with substitution:

From the second equation, we get: $x = v - 3y.$ Substitute this into the first equation: $u = 2(v - 3y) + 4y = 2v - 6y + 4y = 2v - 2y,$ which simplifies to: $y = v - u.$