= Solution
The <characteristic equations for a first-order quasilinear equation> are $\dot x=x$, $\dot y=y$, $\dot u=2u$, with characteristic parameter $s$. Through the prescribed point they give
$$
\boxed{(x(s),y(s),u(s))=(x_0e^s,y_0e^s,u_0e^{2s}).}
$$
For a nonzero projected initial point, the <characteristic projections> are rays from the origin, not trajectories crossing through the origin to the opposite ray. If $x_0=y_0=0$, a classical solution graph requires $u_0=0$, because the original equation reduces there to $0=2u$.
Parameterize the initial line by $(x,y,u)=(a,1-a,a^3)$. Along its characteristic, $x+y=e^s>0$, $a=x/(x+y)$ and $u=a^3e^{2s}$. Therefore
$$
\boxed{u(x,y)=\frac{x^3}{x+y},\qquad x+y>0.}
$$
Differentiation verifies the equation, and substitution of $x+y=1$ verifies the initial data. <Characteristic reachability for radial transport> shows why the domain is this positive half-plane: each point there lies on precisely one characteristic from the data line. Rays in $x+y<0$ never meet that line, so their values are not determined by these data. Allowing a negative scaling factor would incorrectly join two different characteristic rays through a degenerate origin.
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