To find the first integral $\Phi(x, y, C) = 0$ of the differential equation given by $\xi = y = x y' + y' \ln y$ we'll start by analyzing and simplifying the differential equation.

The equation can be rewritten as: $y = x \frac{dy}{dx} + \frac{dy}{dx} \ln y$

Rearranging terms, we get: $y = \frac{dy}{dx} \left( x + \ln y \right)$

Now, solving for $\frac{dy}{dx}$, we obtain: $\frac{dy}{dx} = \frac{y}{x + \ln y}$

Step 1: Rewrite as a Separable Equation

At this stage, let’s try to separate the variables $x$ and $y$.

Rewrite the equation as: $\frac{dy}{y} = \frac{dx}{x + \ln y}$

However, this form is not immediately separable. So, let’s proceed by substituting $v = \ln y$, which implies that $y = e^v$ and $\frac{dy}{dx} = e^v \frac{dv}{dx}$.

Step 2: Substitute and Simplify

Substituting $y = e^v$ and $\frac{dy}{dx} = e^v \frac{dv}{dx}$ into the differential equation, we have: $e^v = x e^v \frac{dv}{dx} + e^v \frac{dv}{dx} \cdot v$ Dividing through by $e^v$, we get: $1 = \frac{dv}{dx} (x + v)$

Thus: $\frac{dv}{dx} = \frac{1}{x + v}$

Step 3: Separate Variables Again

Now we have a separable differential equation in terms of $v$ and $x$: $\int (x + v) \, dv = \int dx$

The next steps involve integrating both sides and solving for $\Phi(x, y, C)$, where $v = \ln y$.

Would you like to continue with these integration steps, or need a specific part clarified further?

---

Here are 5 questions to deepen understanding:

1. How would you approach solving non-linear differential equations?
2. Can you find an integrating factor for this equation?
3. What are the general steps to solve for a first integral?
4. How does substitution help simplify complex differential equations?
5. What alternative methods could solve this type of equation?

Tip: Substitutions are powerful in differential equations, especially when direct separation isn't possible.