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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:
   \[
   y' = \frac{4x + 6 - (4x - 6)}{(2x + 3)^2}
   = \frac{4x + 6 - 4x + 6}{(2x + 3)^2}
   = \frac{12}{(2x + 3)^2}
   \]

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!

Find an equation of the tangent line to the graph of the function y = (2x - 3) / (2x + 3) at the point (3, 1/3).

Learn how to find the equation of the tangent line to the function y = (2x - 3) / (2x + 3) at the point (3, 1/3) using the Quotient Rule. This problem involves applying the derivative and using the point-slope form for the final equation. Suitable for high school calculus students.

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