To implicitly differentiate the equation $y \ln(y) - x \ln(x) = 0$, we follow these steps:

Step 1: Differentiate both sides with respect to $x$.

The given equation is:

$y \ln(y) - x \ln(x) = 0$

We differentiate each term using the product rule, as both $y \ln(y)$ and $x \ln(x)$ involve two functions (in the case of $y \ln(y)$, $y$ is implicitly a function of $x$).

Differentiating $y \ln(y)$:

$\frac{d}{dx} \left( y \ln(y) \right) = \ln(y) \frac{dy}{dx} + \frac{y}{y} \frac{dy}{dx} = \ln(y) \frac{dy}{dx} + \frac{dy}{dx}$

Differentiating $x \ln(x)$:

$\frac{d}{dx} \left( x \ln(x) \right) = \ln(x) + 1$

Step 2: Set up the differentiated equation.

Now we substitute these results into the equation:

$\left( \ln(y) + 1 \right) \frac{dy}{dx} - \left( \ln(x) + 1 \right) = 0$

Step 3: Solve for $\frac{dy}{dx}$.

To solve for $\frac{dy}{dx}$, we rearrange the equation:

$\left( \ln(y) + 1 \right) \frac{dy}{dx} = \ln(x) + 1$

$\frac{dy}{dx} = \frac{\ln(x) + 1}{\ln(y) + 1}$

This is the implicit derivative of $y$ with respect to $x$.

Summary:

The implicit derivative is: $\frac{dy}{dx} = \frac{\ln(x) + 1}{\ln(y) + 1}$

Do you want further details or have any questions on the steps?
Here are 5 related questions to further explore:

1. How does the product rule work in implicit differentiation?
2. What is the chain rule, and how does it apply when differentiating $y \ln(y)$?
3. Can you explain implicit differentiation and how it differs from explicit differentiation?
4. How would you solve $y \ln(y) - x \ln(x) = 0$ for $y$ explicitly?
5. What happens if $y = x$? How would that affect the differentiation?

Tip: When using implicit differentiation, remember to apply the chain rule whenever you differentiate a term involving $y$, as $y$ is a function of $x$.