Finding the uniquely determined region of a PDE

This is Exercise 1.2.7 in Partial Differential Equations by Strauss.

(a) The slope of characteristics is $x/y$. Solve $$ \frac{dy}{dx}=\frac{x}{y} \tag{1} $$ by separating variables: $y\,dy=x\,dx$, hence $y^2/2=x^2/2+C$. Isolate $C$ here: $$y^2-x^2=C \tag{2}$$

The general solution is $u(x,y)=f(y^2-x^2)$, an arbitrary function of the left side of (2). Plugging in the given condition $u(0,y) = e^{-y^2}$, we get $$f(y^2)=e^{-y^2}\tag{3}$$ I prefer to use a new letter, such as $z$, for the argument of $f$. The function $f(z)=e^{-z}$ satisfies (3) and gives $$ {u(x,y)= e^{x^2-y^2}} \tag{4} $$

(b) Looking at (3) more carefully, we see that it only involves the values of $f$ on $[0, \infty)$. Therefore, it is satisfied by any function $f$ such that $f(z)=e^{-z}$ for $z\ge 0$. The values of $f$ for negative $z$ can be arbitrary. It follows that the solution $u(x,y)=f(y^2-x^2)$ is uniquely determined in the region $y^2-x^2\ge 0$ (unpainted on the sketch below) but can be any function of the form $f(y^2-x^2)$ in the region $y^2-x^2 < 0$ (painted).