The problem is asking for the equation of the tangent line to the graph of the function $y = \frac{2x - 3}{2x + 3}$ at the point $\left( 3, \frac{1}{3} \right)$.

Steps to solve:

1. Find the derivative $y'$ (slope of the tangent line) using the quotient rule, which is: $\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$ Here, $f(x) = 2x - 3$ and $g(x) = 2x + 3$.

   Compute:

   • $f'(x) = 2$
   • $g'(x) = 2$

   Applying the quotient rule: $y' = \frac{(2)(2x + 3) - (2x - 3)(2)}{(2x + 3)^2}$ Simplify:

   = \frac{4x + 6 - 4x + 6}{(2x + 3)^2} = \frac{12}{(2x + 3)^2}$$

2. Evaluate the slope at $x = 3$: $y'(3) = \frac{12}{(2(3) + 3)^2} = \frac{12}{(6 + 3)^2} = \frac{12}{9^2} = \frac{12}{81} = \frac{4}{27}$

   The slope of the tangent line at $x = 3$ is $\frac{4}{27}$.

3. Use the point-slope form of the line equation: $y - y_1 = m(x - x_1)$ where $m = \frac{4}{27}$, $x_1 = 3$, and $y_1 = \frac{1}{3}$.

   Substitute: $y - \frac{1}{3} = \frac{4}{27}(x - 3)$

4. Solve for $y$: $y = \frac{4}{27}(x - 3) + \frac{1}{3}$

Thus, the equation of the tangent line is: $y = \frac{4}{27}(x - 3) + \frac{1}{3}$

The correct option is the first one:

$y = \frac{4}{27}(x - 3) + \frac{1}{3}$

Would you like further details on any of these steps?

Here are 5 related questions for practice:

1. How do you use the quotient rule in differentiation?
2. What is the formula for the point-slope form of a line?
3. How do you calculate the derivative of a rational function?
4. Why is the slope of a tangent line equal to the derivative at a point?
5. How can you find the tangent line for more complex functions?

Tip: The quotient rule is key when differentiating functions expressed as one function divided by another!