1. **State the problem:** Solve the differential equation $$\frac{dv}{dx} = \frac{x^2 - xy + y^2}{xy}$$ by substituting $$y = vx$$. 2. **Substitution:** Given $$y = vx$$, then $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$ by the product rule. 3. **Rewrite the original equation:** Substitute $$y = vx$$ into the right side: $$\frac{x^2 - x(vx) + (vx)^2}{x(vx)} = \frac{x^2 - v x^2 + v^2 x^2}{v x^2} = \frac{x^2(1 - v + v^2)}{v x^2} = \frac{1 - v + v^2}{v}$$ 4. **Express $$\frac{dy}{dx}$$ in terms of $$v$$ and $$\frac{dv}{dx}$$:** $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$ 5. **Equate the two expressions for $$\frac{dy}{dx}$$:** $$v + x\frac{dv}{dx} = \frac{1 - v + v^2}{v}$$ 6. **Isolate $$\frac{dv}{dx}$$:** $$x\frac{dv}{dx} = \frac{1 - v + v^2}{v} - v = \frac{1 - v + v^2 - v^2}{v} = \frac{1 - v}{v}$$ 7. **Simplify:** $$\frac{dv}{dx} = \frac{1 - v}{v x}$$ 8. **Separate variables:** $$\frac{v}{1 - v} dv = \frac{1}{x} dx$$ 9. **Integrate both sides:** Left side: $$\int \frac{v}{1 - v} dv = \int \frac{v - 1 + 1}{1 - v} dv = \int \left(-1 + \frac{1}{1 - v}\right) dv = \int -1 dv + \int \frac{1}{1 - v} dv = -v - \ln|1 - v| + C_1$$ Right side: $$\int \frac{1}{x} dx = \ln|x| + C_2$$ 10. **Combine constants and write the implicit solution:** $$-v - \ln|1 - v| = \ln|x| + C$$ 11. **Recall substitution $$v = \frac{y}{x}$$:** $$-\frac{y}{x} - \ln\left|1 - \frac{y}{x}\right| = \ln|x| + C$$ This is the implicit solution to the differential equation. **Final answer:** $$-\frac{y}{x} - \ln\left|1 - \frac{y}{x}\right| = \ln|x| + C$$