Let's solve the expression step by step.

We are given:

$\frac{\left(\frac{2\left(x-y\right)^{2}}{\left(x+y\right)}\cdot \frac{28}{3\cdot \left(y-x\right)}\cdot \left(-\frac{x+y}{8}\right)-\frac{x-y}{3}\right)}{\left(x-y\right)}$

Step 1: Simplify the inner terms

Let's start by simplifying the expression inside the numerator:

$\frac{2(x-y)^2}{x+y} \cdot \frac{28}{3(y-x)} \cdot \left(-\frac{x+y}{8}\right)$

First, observe that $y-x = -(x-y)$, so we can rewrite $\frac{28}{3(y-x)}$ as:

$\frac{28}{3(y-x)} = \frac{28}{3(-(x-y))} = -\frac{28}{3(x-y)}$

So, the expression becomes:

$\frac{2(x-y)^2}{x+y} \cdot -\frac{28}{3(x-y)} \cdot \left(-\frac{x+y}{8}\right)$

Now, multiply the terms:

$-\frac{28 \cdot 2(x-y)^2}{3(x-y)(x+y)} \cdot -\frac{(x+y)}{8}$

Simplify the negatives:

$\frac{56(x-y)^2}{3(x-y)(x+y)} \cdot \frac{(x+y)}{8}$

Step 2: Cancel out common factors

Now, simplify by canceling out the $(x+y)$ terms:

$\frac{56(x-y)^2}{3 \cdot 8(x-y)} = \frac{56(x-y)}{24} = \frac{7(x-y)}{3}$

Step 3: Incorporate the remaining part of the expression in the numerator

Now, substitute back into the original expression:

$\frac{\frac{7(x-y)}{3} - \frac{x-y}{3}}{x-y}$

Simplify the numerator:

$\frac{7(x-y) - (x-y)}{3(x-y)} = \frac{6(x-y)}{3(x-y)} = 2$

Final Answer

$\text{The expression simplifies to } \boxed{2}.$

Would you like further details or have any questions? Here are five related questions that expand on this topic:

1. How do you handle similar algebraic expressions with different variables?
2. What is the significance of canceling terms in rational expressions?
3. Can you simplify an expression that involves multiple fractions with variables?
4. How do you determine when an expression is fully simplified?
5. What are common mistakes to avoid when simplifying complex fractions?

Tip: When simplifying complex algebraic expressions, always look for common factors to cancel out, which can significantly reduce the complexity of the problem.