 Solution

Choose characteristic parameter $t$ with initial value at $t=0$. The characteristic equations areThe first two integrate immediately; substitution into the third givesEliminating $t$ gives the projected parabolaThe sign restriction selects the half-parabola actually reached from the initial point.

The projection map $(s,t)\mapsto (s{e}^t,s{e}^{2t})$ has Jacobian determinant $s{e}^{3t}\ne 0$, so the initial curve is noncharacteristic for every allowed $s$. Its global inverse isdefined precisely when $x,y$ have the same nonzero sign. Every such point is reached by exactly one characteristic. Hence the region of unique determination from these initial data projects ontothe first and third quadrants. The axes are not reached for finite parameter from $s\ne 0$, and the other two quadrants have no method of characteristics issued from this initial curve.

As a direct check, substitution of the inverse parameters gives the solutionIt has $z(s,s)=s^2$, and differentiating gives $x{z}_x+2y{z}_y=xy$. This is quadrant uniqueness for unequal dilation characteristics.