The equation provided in the image is:

$x^y = y^x$

We need to find $y'$ (the derivative of $y$ with respect to $x$).

Steps to solve:

1. Start by taking the natural logarithm of both sides to simplify the powers: $\ln(x^y) = \ln(y^x)$ Using logarithmic properties ($\ln(a^b) = b \ln(a)$): $y \ln(x) = x \ln(y)$

2. Differentiate both sides of this equation with respect to $x$. Apply implicit differentiation:

   For the left-hand side $y \ln(x)$: $\frac{d}{dx} [y \ln(x)] = y' \ln(x) + y \cdot \frac{1}{x}$

   For the right-hand side $x \ln(y)$: $\frac{d}{dx} [x \ln(y)] = \ln(y) + x \cdot \frac{y'}{y}$

3. Now set the two derivatives equal: $y' \ln(x) + \frac{y}{x} = \ln(y) + \frac{x y'}{y}$

4. Collect terms involving $y'$ on one side: $y' \ln(x) - \frac{x y'}{y} = \ln(y) - \frac{y}{x}$

5. Factor out $y'$ on the left-hand side: $y' \left( \ln(x) - \frac{x}{y} \right) = \ln(y) - \frac{y}{x}$

6. Solve for $y'$: $y' = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}$

Thus, the derivative $y'$ is:

$y' = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}$

Would you like further clarification on any step or more details about logarithmic differentiation?

Here are 5 related questions you might explore:

1. How do we handle logarithmic differentiation in other types of equations?
2. What are the general rules for implicit differentiation?
3. Can we apply this method to non-exponential equations?
4. How does the product rule apply in implicit differentiation?
5. What happens when $x = y$ in the original equation?

Tip: Always remember to apply the chain rule when differentiating functions involving products of variables!