` y^(2)-x^(2) (dy)/(dx) = xy(dy)/(dx)`

Answer

To solve the differential equation \( y^2 - x^2 \frac{dy}{dx} = xy \frac{dy}{dx} \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ y^2 - x^2 \frac{dy}{dx} = xy \frac{dy}{dx} \]
We can rearrange this by bringing all terms involving \(\frac{dy}{dx}\) to one side:
\[ y^2 = xy \frac{dy}{dx} + x^2 \frac{dy}{dx} \]
Factoring out \(\frac{dy}{dx}\):
\[ y^2 = \frac{dy}{dx} (xy + x^2) \]

### Step 2: Isolating \(\frac{dy}{dx}\)
Now, we isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{y^2}{xy + x^2} \]

### Step 3: Substituting \(y = ux\)
Next, we use the substitution \(y = ux\), where \(u\) is a function of \(x\). Then, we have:
\[ \frac{dy}{dx} = u + x \frac{du}{dx} \]
Substituting \(y\) and \(\frac{dy}{dx}\) into our equation gives:
\[ u + x \frac{du}{dx} = \frac{(ux)^2}{x(ux) + x^2} \]
This simplifies to:
\[ u + x \frac{du}{dx} = \frac{u^2 x^2}{x^2(u + 1)} \]
Canceling \(x^2\) from both sides:
\[ u + x \frac{du}{dx} = \frac{u^2}{u + 1} \]

### Step 4: Rearranging for Separation of Variables
Rearranging the equation, we get:
\[ x \frac{du}{dx} = \frac{u^2}{u + 1} - u \]
This simplifies to:
\[ x \frac{du}{dx} = \frac{u^2 - u(u + 1)}{u + 1} = \frac{-u}{u + 1} \]

### Step 5: Separating Variables
Now we can separate the variables:
\[ \frac{(u + 1)}{-u} du = \frac{dx}{x} \]

### Step 6: Integrating Both Sides
Integrating both sides:
\[ \int \frac{(u + 1)}{-u} du = \int \frac{dx}{x} \]
The left side can be split:
\[ \int \left(-1 - \frac{1}{u}\right) du = -\ln |x| + C \]
This results in:
\[ -u - \ln |u| = -\ln |x| + C \]

### Step 7: Substituting Back
Since \(u = \frac{y}{x}\), we substitute back:
\[ -\frac{y}{x} - \ln \left|\frac{y}{x}\right| = -\ln |x| + C \]
This can be rearranged to:
\[ \ln |y| + \frac{y}{x} = C + \ln |x| \]

### Final Solution
Thus, the general solution of the differential equation is:
\[ \ln |y| + \frac{y}{x} = C + \ln |x| \]