= Solution
Characteristics satisfy $dx/ds=1$, $dy/ds=x$, $du/ds=0$. From $(\xi,0)$ they are $x=\xi+s$, $y=\xi s+s^2/2$, $u=f(\xi)$, so $x^2-2y=\xi^2$. The initial parameter is recovered on the two initial half-axes by
$$
\boxed{u(x,y)=f\left(\operatorname{sgn}(x)\sqrt{x^2-2y}\right),\qquad x\ne0,\quad y<x^2/2.}
$$
This is $C^1$ on the displayed <open set>, takes the prescribed data at $y=0,x\ne0$, and substitution verifies $u_x+xu_y=0$. Each branch may extend separately beyond this chosen domain, but arbitrary data on both half-axes need not allow one common extension.
The initial line has normal $(0,1)$; the transport field is $(1,x)$, whose normal component is $x$. Thus it is non-characteristic exactly away from the origin. At the origin the equation itself would require $u_x(0,0)=f'(0)=0$. For generic $f$ this fails, so a $C^1$ solution on a neighborhood of the entire initial line is not guaranteed. The parabola $y=x^2/2$ is the envelope where the recovery of $\xi$ degenerates. Characteristics can also meet both initial half-axes, imposing compatibility such as $f(\xi)=f(-\xi)$ on an attempted larger domain.
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