1. **State the problem:** Solve the differential equation $$\left(x + ye^{\frac{y}{x}}\right) dx - xe^{\frac{y}{x}} dy = 0$$ with initial condition $$y(1) = 0.$$\n\n2. **Rewrite the equation:** We have $$\left(x + ye^{\frac{y}{x}}\right) dx = xe^{\frac{y}{x}} dy.$$\nDividing both sides by $dx$, $$x + ye^{\frac{y}{x}} = xe^{\frac{y}{x}} \frac{dy}{dx}.$$\n\n3. **Isolate $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{x + ye^{\frac{y}{x}}}{xe^{\frac{y}{x}}} = \frac{x}{xe^{\frac{y}{x}}} + \frac{ye^{\frac{y}{x}}}{xe^{\frac{y}{x}}} = \frac{1}{e^{\frac{y}{x}}} + \frac{y}{x}.$$\n\n4. **Simplify:** $$\frac{dy}{dx} = e^{-\frac{y}{x}} + \frac{y}{x}.$$\n\n5. **Use substitution:** Let $$v = \frac{y}{x} \implies y = vx.$$\nThen $$\frac{dy}{dx} = v + x\frac{dv}{dx}$$ by the product rule.\n\n6. **Substitute into the differential equation:** $$v + x\frac{dv}{dx} = e^{-v} + v.$$\n\n7. **Simplify:** $$x\frac{dv}{dx} = e^{-v}.$$\n\n8. **Separate variables:** $$\frac{dv}{e^{-v}} = \frac{dx}{x}.$$\nRewrite the left side: $$e^{v} dv = \frac{dx}{x}.$$\n\n9. **Integrate both sides:** $$\int e^{v} dv = \int \frac{dx}{x}.$$\n\n10. **Integrate:** $$e^{v} = \ln|x| + C,$$ where $C$ is the constant of integration.\n\n11. **Back-substitute $v = \frac{y}{x}$:** $$e^{\frac{y}{x}} = \ln|x| + C.$$\n\n12. **Apply initial condition $y(1) = 0$:** Substitute $x=1$, $y=0$:\n$$e^{\frac{0}{1}} = e^0 = 1 = \ln 1 + C = 0 + C \implies C = 1.$$\n\n13. **Final implicit solution:** $$e^{\frac{y}{x}} = \ln|x| + 1.$$\n\nThis is the implicit solution to the differential equation satisfying the initial condition.