1. **State the problem:** Solve the differential equation $$x(x+y)\frac{dy}{dx} - y(3x+y) = 0$$ using the substitution $$y = vx$$. 2. **Substitution:** Let $$y = vx$$, so $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$ by the product rule. 3. **Rewrite the equation:** Substitute $$y = vx$$ and $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$ into the original equation: $$x(x + vx)(v + x\frac{dv}{dx}) - vx(3x + vx) = 0$$ 4. **Simplify terms:** $$x^2(1 + v)(v + x\frac{dv}{dx}) - x^2 v(3 + v) = 0$$ 5. **Divide both sides by $$x^2$$ (assuming $$x \neq 0$$):** $$\cancel{x^2}(1 + v)(v + x\frac{dv}{dx}) - \cancel{x^2} v(3 + v) = 0$$ 6. **Expand:** $$(1 + v)v + (1 + v)x\frac{dv}{dx} - v(3 + v) = 0$$ 7. **Simplify:** $$(1 + v)v - v(3 + v) + (1 + v)x\frac{dv}{dx} = 0$$ Calculate the first two terms: $$(v + v^2) - (3v + v^2) = v + v^2 - 3v - v^2 = -2v$$ So the equation becomes: $$-2v + (1 + v)x\frac{dv}{dx} = 0$$ 8. **Isolate $$\frac{dv}{dx}$$:** $$(1 + v)x\frac{dv}{dx} = 2v$$ 9. **Divide both sides by $$(1 + v)x$$:** $$\frac{dv}{dx} = \frac{2v}{x(1 + v)}$$ 10. **Separate variables:** $$\frac{1 + v}{v} dv = \frac{2}{x} dx$$ Rewrite the left side: $$\left(\frac{1}{v} + 1\right) dv = \frac{2}{x} dx$$ 11. **Integrate both sides:** $$\int \left(\frac{1}{v} + 1\right) dv = \int \frac{2}{x} dx$$ 12. **Perform integration:** $$\int \frac{1}{v} dv + \int 1 dv = 2 \int \frac{1}{x} dx$$ $$\ln|v| + v = 2 \ln|x| + C$$ 13. **Substitute back $$v = \frac{y}{x}$$:** $$\ln\left|\frac{y}{x}\right| + \frac{y}{x} = 2 \ln|x| + C$$ 14. **Rewrite:** $$\ln|y| - \ln|x| + \frac{y}{x} = 2 \ln|x| + C$$ $$\ln|y| + \frac{y}{x} = 3 \ln|x| + C$$ This implicit solution relates $$x$$ and $$y$$. **Final answer:** $$\boxed{\ln|y| + \frac{y}{x} = 3 \ln|x| + C}$$