= Solution
Start a <characteristic curve> at $(0,\eta)$ with $u=-\eta$. The <Characteristic curve> equations give $u=-\eta$, $y=\eta+s$, $x=-\eta s$, so $x=\eta^2-\eta y$. Therefore the two branches are
$$
\boxed{u_\pm(x,y)=-\frac{y\pm\sqrt{y^2+4x}}2.}
$$
Each is $C^1$ on $D=\{y^2+4x>0\}$. The plus branch matches $u(0,y)=-y$ for $y>0$, whereas the minus branch matches it for $y<0$. Differentiating $u^2+yu-x=0$ gives $u_x=1/(2u+y)$ and $u_y=-u/(2u+y)$, verifying $u_y+uu_x=0$ wherever $2u+y\ne0$.
On the initial line, the normal <Characteristic curve> speed is $u=-y$, nonzero away from the origin. Thus the local existence theorem gives precisely these separate local branches near positive and negative initial points. At $y^2+4x=0$ the branches merge and their derivatives become singular. There is no $C^1$ solution realizing the entire initial line in a neighborhood of the origin: the data force $u_y(0,0)=-1$, while the equation and $u(0,0)=0$ force $u_y(0,0)=0$. The failure at this <Characteristic curve> point is a genuine limitation of the printed global initial-data request.
Back to article page
