$$yu_x-xu_y+u_z=1$$
The change to cylindrical coordinates would simplify the calculus. Nevertheless, we will solve the PDE in Cartesian coo1dinates, just to show that it is not very difficult.
The differential characteristic equations are :
$\quad \frac{dx}{y}= \frac{dy}{-x}=\frac{dz}{1}=\frac{du}{1}$
A first family of characteristic curves comes from $dz=du \quad\to\quad u-z=c_1$
A second family of characteristic curves comes from :$\quad\frac{dx}{y}= \frac{dy}{-x} \quad\to\quad x^2+y^2=c_2$
A third family is a bit more difficult to find : $ \frac{dz}{1}=\frac{dx}{\sqrt{c_2-x^2}} \quad\to\quad \tan^{-1}\left(\frac{x}{\sqrt{c_2-x^2}} \right)-z=c_3 $
On the characteristic curves $c_1,c_2,c_3$ are independent. Elsewhere they are related by an implicit equation :
$$\Phi\left(u-z\:,\:x^2+y^2\:,\:\tan^{-1}\left(\frac{x}{y} \right)-z\right)=0$$
where $\Phi$ is any differentiable function of three variables.
An equivalant manner to express the relationship is :
$$u-z=F\left(x^2+y^2\:,\:\tan^{-1}\left(\frac{x}{y} \right)-z\right)$$
where $F(X,Y)$ is any differentiable function of two variables.
This is the general solution of the PDE in Cartesian coordinates :
$$u(x,y,z)=z+F\left(x^2+y^2\:,\:\tan^{-1}\left(\frac{x}{y} \right)-z\right)$$
Or in cylindrical coordinates :
$$u=z+F\left(\rho^2\:,\:\tan^{-1}\left(\frac{\cos(\theta)}{\sin(\theta)} \right)-z\right) =z+F\left(\rho^2\:,\:\frac{\pi}{2}-\theta-z\right) $$
Now, we consider the boundary condition :
$$u(x,y,0)=x+y=\rho\left(\cos(\theta)+\sin(\theta)\right)=F\left(\rho^2\:,\:\frac{\pi}{2}-\theta\right)$$
Thus, the function $F(X,Y)$ is determined :
$$F(X,Y)=\sqrt{X}\left(\cos(\frac{\pi}{2}-Y)+\sin(\frac{\pi}{2}-Y) \right)=\sqrt{X}\left(\sin(Y)+\cos(Y) \right)$$
Bringing it back into the above general solution, with $X=\sqrt{x^2+y^2}$ and $Y=\tan^{-1}\left(\frac{x}{y} \right)-z$ , the particular solution of the PDE according to the boundary condition is :
$$u(x,y,z)=z+\sqrt{x^2+y^2}\left[\sin\left(\tan^{-1}\left(\frac{x}{y} \right)-z \right) +\cos \left(\tan^{-1}\left(\frac{x}{y} \right)-z \right) \right]$$
First family: “This allows me to divide both sides by $x^2$.” No, you can't. The variables $x$ and $t$ are not independent, so you can't take $x^2$ outside of the $t$ integral.
What you have is
$$dt/t^2= dx/x^2 \iff d(-1/t)=d(-1/x) \iff d(1/t-1/x)=0 \iff 1/t-1/x=c_1.$$
Second family: If $A=B$, then $(tA+xB)/(t+x)=(tA+xA)/(t+x)=A(t+x)/(t+x)=A$ (which also equals $B$). That is,
$$
A=B
\quad\implies\quad
A = B = \frac{tA+xB}{t+x}
.
$$
The computation starts out by applying this identity with $A=dt/t^2$ and $B=dx/x^2$.
And then you get
$$
dt/t + dx/x = du/u
\iff
d \ln|t| + d \ln|x| = d \ln|u|
\iff
d \ln|\tfrac{u}{xt}|=0
\iff
\tfrac{u}{xt}=c_2
.
$$
(If the logarithm of the absolute value of something is constant, then that something itself is constant too.)
Best Answer
Take a point on the curve $x=y^2$, say $(x,y)=(a^2,a)$. It is given that $u=1$ at such a point. We can assume that the characteristic curve through the point $(x,y,u)=(a^2,a,1)$ passes it when $s=0$. So you can let $x(0)=a^2$, $y(0)=a$ and $u(0)=1$ to solve for $c_x$, $c_y$ and $c_u$. Once this is done, you know $x$, $y$ and $u$ as functions of $a$ and $s$. Now take the expressions for $x$ and $y$ and try to solve for $a$ and $s$, and then insert the resulting expressions for $a=a(x,y)$ and $s=s(x,y)$ into the expression for $u$, to obtain the solution $u(x,y)$.