1. **State the problem:** Solve the differential equation $ (x + y e^{\frac{y}{x}}) \, dx - x e^{\frac{y}{x}} \, dy = 0 $ with initial condition $ y(1) = 0 $.\n\n2. **Rewrite the equation:** The given equation can be written as \n$$ (x + y e^{\frac{y}{x}}) + (- x e^{\frac{y}{x}}) \frac{dy}{dx} = 0 $$\nwhich implies\n$$ \frac{dy}{dx} = \frac{x + y e^{\frac{y}{x}}}{x e^{\frac{y}{x}}} $$\n\n3. **Simplify the right side:**\n$$ \frac{dy}{dx} = \frac{x}{x e^{\frac{y}{x}}} + \frac{y e^{\frac{y}{x}}}{x e^{\frac{y}{x}}} = e^{-\frac{y}{x}} + \frac{y}{x} $$\n\n4. **Substitute:** Let $ v = \frac{y}{x} \implies y = vx $. Then\n$$ \frac{dy}{dx} = v + x \frac{dv}{dx} $$\n\n5. **Replace $\frac{dy}{dx}$ in the equation:**\n$$ v + x \frac{dv}{dx} = e^{-v} + v $$\n\n6. **Simplify:**\n$$ x \frac{dv}{dx} = e^{-v} $$\n\n7. **Separate variables:**\n$$ \frac{dv}{e^{-v}} = \frac{dx}{x} $$\nwhich is\n$$ e^{v} dv = \frac{dx}{x} $$\n\n8. **Integrate both sides:**\n$$ \int e^{v} dv = \int \frac{dx}{x} $$\n$$ e^{v} = \ln|x| + C $$\n\n9. **Rewrite in terms of $y$ and $x$:**\n$$ e^{\frac{y}{x}} = \ln|x| + C $$\n\n10. **Apply initial condition $y(1) = 0$:**\n$$ e^{\frac{0}{1}} = \ln|1| + C \Rightarrow 1 = 0 + C \Rightarrow C = 1 $$\n\n11. **Final solution:**\n$$ \ln|x| = e^{\frac{y}{x}} - 1 $$\n\n**Answer:** Option D\n