Let's solve the expression shown in the image. The expression given is:

$\left(\frac{3x^{\frac{3}{2}} y^3}{x^2 y^{-\frac{1}{2}}}\right)^{-2}$

Step 1: Simplify the expression inside the parentheses

We start by simplifying the fraction inside the parentheses:

$\frac{3x^{\frac{3}{2}} y^3}{x^2 y^{-\frac{1}{2}}}$

This can be split into separate parts for $x$ and $y$:

$3 \cdot \frac{x^{\frac{3}{2}}}{x^2} \cdot \frac{y^3}{y^{-\frac{1}{2}}}$

For $x$:

$\frac{x^{\frac{3}{2}}}{x^2} = x^{\frac{3}{2} - 2} = x^{\frac{3}{2} - \frac{4}{2}} = x^{-\frac{1}{2}}$

For $y$:

$\frac{y^3}{y^{-\frac{1}{2}}} = y^{3 - \left(-\frac{1}{2}\right)} = y^{3 + \frac{1}{2}} = y^{\frac{7}{2}}$

So, the simplified expression inside the parentheses becomes:

$\frac{3x^{-\frac{1}{2}} y^{\frac{7}{2}}}{1} = 3x^{-\frac{1}{2}} y^{\frac{7}{2}}$

Step 2: Apply the outer exponent $-2$

Now, apply the exponent $-2$ to the entire expression:

$\left(3x^{-\frac{1}{2}} y^{\frac{7}{2}}\right)^{-2}$

Distribute the exponent $-2$ to each term:

$3^{-2} \cdot \left(x^{-\frac{1}{2}}\right)^{-2} \cdot \left(y^{\frac{7}{2}}\right)^{-2}$

This simplifies to:

$\frac{1}{9} \cdot x^{1} \cdot y^{-7}$

Or more simply:

$\frac{x}{9y^7}$

Final Answer:

The simplified expression without negative exponents is:

$\boxed{\frac{x}{9y^7}}$

Would you like any additional details or have any questions about this solution?

Here are five related questions to expand your understanding:

1. How would the solution change if the outer exponent were positive?
2. What happens if we introduce different fractional exponents for $x$ and $y$?
3. How do you handle negative exponents when they are in the denominator?
4. How does the base change affect the exponentiation process?
5. How do you simplify an expression with multiple variables and different exponents?

Tip: When dealing with exponents, always apply exponent rules systematically to avoid mistakes, especially with negative exponents.