$$y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y}=0$$
System of characteristic ODEs :
$$\frac{\mathrm{d}x}{y} = \frac{\mathrm{d}y}{-x} = \frac{\mathrm{d}z}{0}$$
Take care that it isn't $\frac{\mathrm{d}y}{x}$ but $\frac{\mathrm{d}y}{-x}$
A first family of characteristic curves comes from $\frac{\mathrm{d}z}{0}=$finite $\implies \mathrm{d}z=0 \implies z=c_1$.
A second family of characteristic curves comes from $\frac{\mathrm{d}x}{y} = -\frac{\mathrm{d}y}{x}\implies x^2+y^2=c_2$.
The general solution of the PDE expressed on the form of implicit equation is :
$$\Phi(z\:,\:x^2+y^2)=0$$
where $\Phi$ is an arbitrary function of two variables.
Or equivalently on explicit form :
$$z=F(x^2+y^2)$$
where $F$ is an arbitrary function.
The arbitrary function has to be determined according to the boundary condition.
In the wording of the question, it is said that the boundary is a curve which equation is $x^2+y^2=r^2$, that is a circle. So the boundary is defined, but the condition is not given :
Citation : 2. Particular solution if $z$ is specified along a circle
$x^2 + y^2 = r^2$ in the $(x,y)-$plane.
What exactly is the specification ?
IN ADDITION :
In comments, the condition is specified as : $z(x,1)=e^x$ with $x>0$.
Note that the condition is not along a circle $x^2+y^2=r^2$ but is along the straight line $y=1$.
Putting this condition into the general solution leads to :
$$F(x^2+1)=e^x$$
Let : $\quad X=x^2+1 \quad\to\quad x=\sqrt{X-1}$
$$F(X)=e^{\sqrt{X-1}}$$
So, the function $F$ is determined. We put it into the above general solution where $X=x^2+y^2$. The particular solution fitting the boundary condition is :
$$z(x,y)=\exp\left(\sqrt{x^2+y^2-1}\right)$$
