Step 1: Rewrite the equation as $\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}$

Step 2: Show it is homogeneous by dividing numerator and denominator by $x^2$: $\frac{dy}{dx} = \frac{1 + \left(\frac{y}{x}\right)^2}{1 + \frac{y}{x}}$, which is of the form $f\left(\frac{y}{x}\right)$

Step 3: Substitute $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$

Step 4: This gives $v + x\frac{dv}{dx} = \frac{1 + v^2}{1 + v}$, simplifying to $x\frac{dv}{dx} = \frac{1 - v}{1 + v}$

Step 5: Separate variables: $\frac{1+v}{1-v}dv = \frac{dx}{x}$

Step 6: Simplify $\frac{1+v}{1-v} = -1 + \frac{2}{1-v}$ and integrate both sides

Step 7: Integration gives $-v - 2\ln|1-v| = \ln|x| + C$

Step 8: Substitute back $v = \frac{y}{x}$ to get $-\frac{y}{x} - 2\ln\left|1-\frac{y}{x}\right| = \ln|x| + C$

Step 9: Simplify to final form: $\ln|x| - 2\ln|x-y| - \frac{y}{x} = C$