Let's simplify the expression step by step:

$\frac{\left( (2x^2 \cdot y)^2 \cdot 5x \cdot y \right)}{\left(3x \cdot y^3\right)^3}$

Step 1: Expand the powers in the numerator

We first expand $(2x^2 \cdot y)^2$. Using the power of a product rule, $(ab)^2 = a^2 \cdot b^2$, we get: $(2x^2 \cdot y)^2 = 2^2 \cdot (x^2)^2 \cdot y^2 = 4x^4y^2$

Now, substitute this back into the numerator: $\text{Numerator} = (4x^4y^2) \cdot 5x \cdot y = 4 \cdot 5 \cdot x^4 \cdot x \cdot y^2 \cdot y = 20x^5y^3$

So the numerator becomes: $20x^5y^3$

Step 2: Expand the denominator

Now expand the denominator $(3x \cdot y^3)^3$. Again, use the power of a product rule: $(3x \cdot y^3)^3 = 3^3 \cdot x^3 \cdot (y^3)^3 = 27x^3y^9$

So the denominator becomes: $27x^3y^9$

Step 3: Simplify the fraction

Now, simplify the expression: $\frac{20x^5y^3}{27x^3y^9}$

For the $x$-terms: $\frac{x^5}{x^3} = x^{5-3} = x^2$

For the $y$-terms: $\frac{y^3}{y^9} = y^{3-9} = y^{-6} = \frac{1}{y^6}$

Final expression

Now combine everything: $\frac{20x^5y^3}{27x^3y^9} = \frac{20x^2}{27y^6}$

Thus, the simplified expression is: $\boxed{\frac{20x^2}{27y^6}}$

Would you like further details or clarifications?

Here are 5 questions to explore:

1. How do you simplify expressions with powers in both the numerator and denominator?
2. What is the power of a product rule, and how is it applied?
3. How does exponent subtraction work when simplifying variables?
4. How do negative exponents impact the final expression?
5. What would happen if the denominator had a higher degree for $x$?

Tip: Always apply exponent rules carefully when simplifying variables.