The given differential equation is $\frac{dy}{dx} = \frac{x+y}{x}$.

This can be rewritten as $\frac{dy}{dx} = 1 + \frac{y}{x}$.

Let $y = vx$, so $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

Substituting into the differential equation, we get $v + x\frac{dv}{dx} = 1 + v$.

This simplifies to $x\frac{dv}{dx} = 1$, so $\frac{dv}{dx} = \frac{1}{x}$.

Integrating both sides with respect to $x$, we get $\int dv = \int \frac{1}{x} dx$, which gives $v = \ln|x| + C$.

Substituting back $v = \frac{y}{x}$, we have $\frac{y}{x} = \ln|x| + C$.

Thus, $y = x\ln|x| + Cx$.

Now, we use the initial condition $y(1) = 0$.

Substituting $x = 1$ and $y = 0$, we get $0 = 1\ln|1| + C(1)$, which simplifies to $0 = 0 + C$, so $C = 0$.

Therefore, the particular solution is $y = x\ln|x|$.