To find the coefficient of \( x^2 y^6 \) in the expansion of \( \left( \frac{3}{5}x - \frac{y}{2} \right)^8 \), we will use the Binomial Theorem. The Binomial Theorem states that:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In our case, we can identify \( a = \frac{3}{5}x \) and \( b = -\frac{y}{2} \), and \( n = 8 \).

### Step 1: Analyze the problem requirements
We need to find the specific term in the expansion that corresponds to \( x^2 y^6 \). This means we need to determine the values of \( k \) and \( n-k \) such that:

- The power of \( x \) is 2.
- The power of \( y \) is 6.

### Step 2: List the steps to solve the problem
1. Identify the general term in the expansion using the Binomial Theorem.
2. Set up the equation for the powers of \( x \) and \( y \).
3. Solve for \( k \) and \( n-k \).
4. Substitute \( k \) into the general term to find the coefficient.
5. Simplify the coefficient to express it as a common fraction.

### Step 3: Execute the solution process
1. The general term in the expansion is given by:

\[
T_k = \binom{8}{k} \left( \frac{3}{5}x \right)^{8-k} \left( -\frac{y}{2} \right)^k
\]

2. We need \( (8-k) \) to equal 2 (for \( x^2 \)) and \( k \) to equal 6 (for \( y^6 \)). Thus, we have:

\[
8 - k = 2 \implies k = 6
\]

3. Now we can substitute \( k = 6 \) into the general term:

\[
T_6 = \binom{8}{6} \left( \frac{3}{5}x \right)^{2} \left( -\frac{y}{2} \right)^{6}
\]

4. Calculate \( \binom{8}{6} \):

\[
\binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28
\]

5. Now substitute \( k = 6 \) into the term:

\[
T_6 = 28 \left( \frac{3}{5}x \right)^{2} \left( -\frac{y}{2} \right)^{6}
\]

Calculating \( \left( \frac{3}{5}x \right)^{2} \):

\[
\left( \frac{3}{5}x \right)^{2} = \frac{9}{25} x^2
\]

Calculating \( \left( -\frac{y}{2} \right)^{6} \):

\[
\left( -\frac{y}{2} \right)^{6} = \left( \frac{y}{2} \right)^{6} = \frac{y^6}{64}
\]

6. Now combine these results:

\[
T_6 = 28 \cdot \frac{9}{25} x^2 \cdot \frac{y^6}{64}
\]

7. Simplifying the coefficient:

\[
T_6 = 28 \cdot \frac{9}{25} \cdot \frac{1}{64} x^2 y^6 = \frac{28 \cdot 9}{25 \cdot 64} x^2 y^6
\]

Calculating \( 28 \cdot 9 \):

\[
28 \cdot 9 = 252
\]

Thus, we have:

\[
T_6 = \frac{252}{1600} x^2 y^6
\]

8. Finally, simplify \( \frac{252}{1600} \):

The GCD of 252 and 1600 is 4, so:

\[
\frac{252 \div 4}{1600 \div 4} = \frac{63}{400}
\]

### Step 4: Provide the final answer
The coefficient of \( x^2 y^6 \) in the expansion of \( \left( \frac{3}{5}x - \frac{y}{2} \right)^8 \) is:

\[
\boxed{\frac{63}{400}}
\]