3D Cauchy problem for the PDE $ yu_x-xu_y+u_z=0 $ I will answer the question myself but let me know what you think of my correctness.
We have the Cauchy Problem 
$$ yu_x-xu_y+u_z=0 $$
 with data $u(x,y,0) = x+y$.
 A: If we parametrize our curve, we have 
$$ x'(t)=y , y'(t)=-x, z'(t)=1 , u'(t)=0 $$
which gives solutions 
$$ x(t)=A\cos(t+\phi) , y(t)=A\sin(t+\phi), z(t)=t+z_0 , u(t)=u_0 $$
From the data we see that $z_0=0$ and
$$u_0= x(0)+y(0)= A(\cos(\phi)+\sin(\phi)) $$
So we have the following 
$$\frac{y}{x} =\tan(t+\phi)  \implies \phi=\arctan(x^{-1}y)-t  $$
$$ x^2+y^2 = A^2(sin^2(t+\phi)+cos^2(t+\phi))= A^2 \implies A=\pm \sqrt{x^2+y^2}$$
$$ z=t+z_0 \implies t=z $$
Then 
$$ u(x,y,z)=A(\cos(\phi)+\sin(\phi))$$
becomes 
$$ u(x,y,z)= \sqrt{x^2+y^2}(\cos(\arctan(x^{-1}y)-z)+\sin(\arctan(x^{-1}y)-z))$$
We take the positive solution but I am not sure how to know that apriori.  
A: The characteristic ODE are
$\dfrac{d x}{ d t} = y$, $\dfrac{d y}{ d t} = - x$, $\dfrac{d z}{ d t} = 1$, $\dfrac{d u}{ d t} = 0$,
with initial conditions $x(0)=x_0$, $y(0)=y_0$, $z(0)=0$, $u(0)=u_0$ at $t=0$.
The solutions of the ODE are
$x = x_0 \cos(t) + y_0 \sin(t)$, $y = - x_0 \sin(t) + y_0 \cos(t) $, $z = t$, $u=u_0$.
By substituting $z$ into $x,y$, there are $x = x_0 \cos(z) + y_0 \sin(z)$, $y = - x_0 \sin(z) + y_0 \cos(z) $.
The constants $x_0,y_0$ are obtained from $x,y$, which are
$x_0 = x \cos(z) - y \sin(z)$, $y_0 = x \sin(z) + y \cos(z)$.
Substitute $x_0,y_0$ into $u_0$,  $u(x,y,z) = u_0(x_0,y_0,0)= x_0 + y_0 = x \cos(z) - y \sin(z) + x \sin(z) + y \cos(z) = (x+y) \cos(z) +(x-y) \sin(z)$.
Thus, the solution of the PDE is obtained, which is
$u(x,y,z)  = (x+y) \cos(z) +(x-y) \sin(z)$.
