How to show that the operator \begin{align*} L[u]\equiv \frac {\partial^2 u}{\partial x^2}+\frac {\partial^2 u}{\partial y^2}-\frac {\partial^2 u}{\partial z^2} \end{align*} is not elliptic?
Define $L[u]\equiv A\frac {\partial^2 u}{\partial x^2}+B\frac {\partial^2 u}{\partial y^2}+C\frac {\partial^2 u}{\partial z^2}+D\frac {\partial^2 u}{\partial x \partial y}+E\frac {\partial^2 u}{\partial x\partial z}+F\frac {\partial^2 u}{\partial y \partial z}+G\frac {\partial u}{\partial x}+H\frac {\partial u}{\partial y}+J\frac {\partial u}{\partial z}+Ku$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $J$ may depend upon $x$, $y$, and $z$ to be elliptic if there is a linear transformation to new coordinates $(\xi, \eta, \zeta)$ such that in these coordinates, \begin{align*} L[u]\equiv \alpha[\frac {\partial^2 u}{\partial \xi^2}+\frac {\partial^2 u}{\partial \eta^2}+\frac {\partial^2 u}{\partial \zeta^2}]+\beta\frac {\partial u}{\partial \xi}+\gamma\frac {\partial u}{\partial \eta}+\delta\frac {\partial u}{\partial \zeta}+Ku \end{align*}
A linear differential operator with constant coefficients is elliptic iff it's characteristic polynomial is non-zero for non-zero inputs.
The characteristic polynomial of your operator is $p(x,y,z)=x^2+y^2-z^2$ and the equation $p(x,y,z)=0$ has plenty of non-zero solutions.