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We are tasked with finding the value of \(\frac{dy}{dx}\) for the equation \(x^y \cdot y^x = 16\) at the point \((2, 2)\).

### Steps:
1. **Given equation**: \(x^y \cdot y^x = 16\)

   We need to differentiate this equation implicitly with respect to \(x\).

2. **Differentiate both sides**:
   Let's first apply the product rule on the left-hand side:
   \[
   \frac{d}{dx}\left(x^y \cdot y^x\right) = \frac{d}{dx}(16)
   \]
   The right-hand side is simply 0 because the derivative of a constant (16) is 0.

3. **Product rule application**: 
   Using the product rule \(\frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v'\), where:
   - \( u = x^y \)
   - \( v = y^x \)
   
   We need to differentiate both \(x^y\) and \(y^x\):
   - For \(x^y\), use the logarithmic differentiation method. The derivative is:
     \[
     \frac{d}{dx}(x^y) = x^y \cdot \left(\ln(x) \cdot \frac{dy}{dx} + \frac{y}{x}\right)
     \]
   - For \(y^x\), again use logarithmic differentiation. The derivative is:
     \[
     \frac{d}{dx}(y^x) = y^x \cdot \left(\ln(y) + \frac{x}{y} \cdot \frac{dy}{dx}\right)
     \]

4. **Substitute in product rule**:
   Substitute the derivatives back into the product rule:
   \[
   \left( x^y \cdot \left(\ln(x) \cdot \frac{dy}{dx} + \frac{y}{x} \right) \right) \cdot y^x + x^y \cdot \left( y^x \cdot \left( \ln(y) + \frac{x}{y} \cdot \frac{dy}{dx} \right) \right) = 0
   \]

5. **Substitute \((x, y) = (2, 2)\)**:
   Now, substitute \(x = 2\) and \(y = 2\) into the equation and solve for \(\frac{dy}{dx}\):
   - \(x^y = 2^2 = 4\)
   - \(y^x = 2^2 = 4\)
   - \(\ln(2) \approx 0.693\)

   The equation simplifies to:
   \[
   \left(4 \cdot \left( 0.693 \cdot \frac{dy}{dx} + \frac{2}{2} \right) \right) \cdot 4 + 4 \cdot \left( 4 \cdot \left( 0.693 + \frac{2}{2} \cdot \frac{dy}{dx} \right) \right) = 0
   \]

6. **Solve for \(\frac{dy}{dx}\)**:
   After simplifying the terms, solving for \(\frac{dy}{dx}\) gives:
   \[
   \frac{dy}{dx} = -1
   \]

### Final Answer: 
\[
\boxed{-1}
\]

Would you like a more detailed breakdown of any specific step? Here are 5 related questions to explore further:

1. How does logarithmic differentiation work?
2. Why is the product rule used in this problem?
3. How do you find the derivative of expressions with mixed powers like \(x^y\)?
4. What is implicit differentiation, and why is it useful here?
5. Can this problem be solved using another approach?

**Tip:** When differentiating functions with variables in both the base and the exponent, logarithmic differentiation can simplify the process significantly.

If x^y * y^x = 16, find the value of dy/dx at the point (2,2).

Learn how to solve the problem x^y * y^x = 16 by using implicit differentiation and logarithmic differentiation. This step-by-step guide shows how to apply the product rule and calculate dy/dx at the point (2, 2). Suitable for Grade 12 or college-level calculus students.

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