Solving a PDE first order in three variables 
Solve the following PDE
$$ u_{x_1} + u_{x_2} + x_3 u_{x_3} = u^3 $$
with cauchy data $u(x_1,x_2,1) = \phi(x_1,x_2) $. Also, determine the values of $x_1,x_2$ and $x_3$ for which the IVP exists.

Try:
The characteristics are given by
$$ \begin{align*}
     \frac{dx_1}{ds} &= 1, \; \; \; x(r_1,r_2,0) = r_1 \\
     \frac{dx_2}{ds} &= 1, \; \; \; x(r_1,r_2,0) = r_2 \\
    \frac{dx_3}{ds} &= x_3, \; \; \; x(r_1,r_2,0) = 1 \\
    \frac{d z}{ds} &= z^3, \; \; \; x(r_1,r_2,0) = h(r_1,r_2) \\
\end{align*}$$
We see easily that $x_1(r_1,r_2,s) = s+r_1$ and similarly $x_2 = s + r_2$ and $x_3 = e^s $. Next, we have
$$ \frac{dz}{z^3} = ds \implies - \frac{1}{2 z^2} = s + C \implies -\frac{1}{2h^2} = C $$
Therefore,
$$ z(r_1,r_2,s) = \frac{ 2 |h(r_1,r_2)| }{\sqrt{1 - 2h(r_1,r_2)} } $$
also, notice that $s = \log x_3 $ and so $r_1 = x_1 - \log x_3$ and $r_2 = x_2 - \log x_3$. Therefore, our solution is
$$ u(x_1,x_2,x_3) = \frac{ 2 | h( x_1 - \log x_3, x_2 - \log x_3) |}{\sqrt{1-2h( x_1 - \log x_3, x_2 - \log x_3)}} $$
and the solution exists for all $(x_1,x_2,x_3)$ since
$$ Jac(x_1,x_2,x_3) = 1 \neq 0 $$
Is this a correct solution?
 A: You have the right idea; you've just wrong somewhere along the way in forgetting a square and dropping an $s$ it looks like. You can always check if your result is a solution or not by directly applying the PDE to it and checking the condition(s).
\begin{cases}
u_x + u_y + z u_z = u^3 \\
u(x, y, 1) = \phi(x, y).
\end{cases}
We have the characteristic
\begin{cases}
\begin{align}
\displaystyle &\frac{dx}{dt} = 1 &x(0) &= s_1 &\implies x(t) &= s_1 + t \\
\displaystyle &\frac{dy}{dt} = 1 &y(0) &= s_2 &\implies y(t) &= s_2 + t \\
\displaystyle &\frac{dz}{dt} = z &z(0) &= 1 &\implies z(t) &= e^t \\
\displaystyle &\frac{d\tilde{u}}{dt} = \tilde{u}^3  &\tilde{u}(0) &= \phi\big( s_1, s_2 \big) &\implies \tilde{u}(t) &= \sqrt{\frac{\phi^2(s_1, s_2)}{1-2\phi^2(s_1, s_2) t}}.
\end{align}
\end{cases}
The system is readily solvable for $\{t, s_1, s_2\}.$
$$u(x,y,z) = \sqrt{\frac{\phi^2(x - \log z, y - \log z)}{1 - 2\phi^2(x - \log z, y - \log z)\log z}}.$$
For the Jacobian, I get
$$J = z.$$
However, unless using the complex-valued logarithm, I would restrict $z$ to be positive.
