Given that $y = \frac{3\text{e}^{2x}}{2x+1}$, find the value of $k$ for which $\frac{\text{d}y}{\text{d}x} = \frac{kxy}{2x+1}$.

1. Differentiate $y$ with respect to $x$ using the quotient rule.

Let $u = 3\text{e}^{2x}$ and $v = 2x+1$.
$\frac{\text{d}u}{\text{d}x} = 6\text{e}^{2x}$
$\frac{\text{d}v}{\text{d}x} = 2$

$\frac{\text{d}y}{\text{d}x} = \frac{v\frac{\text{d}u}{\text{d}x} - u\frac{\text{d}v}{\text{d}x}}{v^2}$
$\frac{\text{d}y}{\text{d}x} = \frac{(2x+1)(6\text{e}^{2x}) - (3\text{e}^{2x})(2)}{(2x+1)^2}$

2. Simplify the numerator of the derivative.

$\frac{\text{d}y}{\text{d}x} = \frac{12x\text{e}^{2x} + 6\text{e}^{2x} - 6\text{e}^{2x}}{(2x+1)^2}$
$\frac{\text{d}y}{\text{d}x} = \frac{12x\text{e}^{2x}}{(2x+1)^2}$

3. Express the derivative in the given form to find $k$.

We are given $\frac{\text{d}y}{\text{d}x} = \frac{kxy}{2x+1}$.
Substitute $y = \frac{3\text{e}^{2x}}{2x+1}$ into this expression:
$\frac{\text{d}y}{\text{d}x} = \frac{kx}{2x+1} \left( \frac{3\text{e}^{2x}}{2x+1} \right) = \frac{3kx\text{e}^{2x}}{(2x+1)^2}$

Equating the two expressions for $\frac{\text{d}y}{\text{d}x}$:
$\frac{12x\text{e}^{2x}}{(2x+1)^2} = \frac{3kx\text{e}^{2x}}{(2x+1)^2}$

Comparing the coefficients of $x\text{e}^{2x}$:
$12 = 3k$
$k = 4$