Given the differential equation: $xdy - ydx = \sqrt{x^2 + y^2} dx$

Divide by $x^2$: $\frac{xdy - ydx}{x^2} = \frac{\sqrt{x^2 + y^2}}{x^2} dx$

Rewrite the left side as $d(\frac{y}{x})$: $d(\frac{y}{x}) = \sqrt{\frac{x^2 + y^2}{x^4}} dx = \sqrt{\frac{1}{x^2} + \frac{y^2}{x^4}} dx = \sqrt{\frac{1}{x^2} + (\frac{y}{x^2})^2} dx$

Let $v = \frac{y}{x}$, then $y = vx$. So, $d(\frac{y}{x}) = \frac{1}{x}\sqrt{1 + v^2} dx$

Thus, $dv = \frac{1}{x}\sqrt{1 + v^2} dx$, which implies $\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$

Integrate both sides: $\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$

$\sinh^{-1}(v) = \ln|x| + C$

Substitute back $v = \frac{y}{x}$: $\sinh^{-1}(\frac{y}{x}) = \ln|x| + C$

Apply the initial condition $y(1) = 0$: $\sinh^{-1}(\frac{0}{1}) = \ln|1| + C$, which gives $0 = 0 + C$, so $C = 0$

Therefore, $\sinh^{-1}(\frac{y}{x}) = \ln|x|$

$\frac{y}{x} = \sinh(\ln|x|)$

We want to find $y(3)$, so $x = 3$: $y(3) = 3\sinh(\ln 3)$

Recall that $\sinh(u) = \frac{e^u - e^{-u}}{2}$, so $\sinh(\ln 3) = \frac{e^{\ln 3} - e^{-\ln 3}}{2} = \frac{3 - \frac{1}{3}}{2} = \frac{\frac{8}{3}}{2} = \frac{4}{3}$

Thus, $y(3) = 3 \cdot \frac{4}{3} = 4$