1. **State the problem:** Rewrite the implicit solution $$-\frac{y}{x} - \ln\left|1 - \frac{y}{x}\right| = \ln|x| + C$$ in the form $(x - y) e^{y/x} = d$, where $d$ is a constant. 2. **Start from the implicit solution:** $$-\frac{y}{x} - \ln\left|1 - \frac{y}{x}\right| = \ln|x| + C$$ 3. **Multiply both sides by $-1$:** $$\frac{y}{x} + \ln\left|1 - \frac{y}{x}\right| = -\ln|x| - C$$ 4. **Rewrite the right side as a single logarithm:** Let $D = e^{-C}$ (absorbing the constant), then $$\frac{y}{x} + \ln\left|1 - \frac{y}{x}\right| = \ln\left(\frac{D}{x}\right)$$ 5. **Exponentiate both sides:** $$e^{\frac{y}{x} + \ln|1 - \frac{y}{x}|} = e^{\ln(\frac{D}{x})}$$ 6. **Use properties of exponents and logarithms:** $$e^{\frac{y}{x}} \cdot \left|1 - \frac{y}{x}\right| = \frac{D}{x}$$ 7. **Multiply both sides by $x$:** $$x e^{\frac{y}{x}} \left|1 - \frac{y}{x}\right| = D$$ 8. **Rewrite $1 - \frac{y}{x} = \frac{x - y}{x}$:** $$x e^{\frac{y}{x}} \left|\frac{x - y}{x}\right| = D$$ 9. **Simplify:** $$x e^{\frac{y}{x}} \frac{|x - y|}{x} = D \implies |x - y| e^{\frac{y}{x}} = D$$ 10. **Since $D$ is an arbitrary positive constant, absorb the absolute value into $d$:** $$ (x - y) e^{\frac{y}{x}} = d $$ where $d = \pm D$ is an arbitrary constant. **Final answer:** $$ (x - y) e^{\frac{y}{x}} = d $$