The Cauchy-Kowalevsky theorem is stated in my notes as:
For the Cauchy problem: $$ \begin{cases} u_{y}=F(x,y,u,u_{x}) \\ u(x,0)=h(x) \end{cases} $$ If $h$ is analytic in a neighborhood of $0$ and $F$ is analytic in a neighborhood of $F(0,0,h(0),h'(0))$, then there exists a unique analytic solution for the PDE in a neighborhood of $(0,0)$.
Well, so suppose I have the problem:
$$ \begin{cases} xu_{x}+yu_{y}+u_{z}=u \\ u(x,y,0)=h(x,y) \end{cases}$$
We have $u_{y}=\frac{u-xu_{x}-u_{z}}{y}=F(x,y,u,u_{x})$. As $F$ is not analytic in $(0,0,h(0),h'(0))$, we cannot apply Cauchy-Kowalevsky theorem, can we?