Thanks Evinda, I can solve in this form...
As $y(x)=-2x+C$ is the characteristic curve, we have that $y+2x=C.$ Let $\xi(x,y)=y+2x$ and $\eta=x$, it follow that:
$\dfrac{\partial \xi}{\partial x}=2, \quad \dfrac{\partial \eta}{\partial x}=1.$
$\dfrac{\partial \xi}{\partial y}=1, \quad \dfrac{\partial \eta}{\partial y}=0.$
Then,
$\dfrac{\partial}{\partial x}=\dfrac{\partial}{\partial \xi}\dfrac{\partial \xi}{\partial x}+\dfrac{\partial}{\partial \eta}\dfrac{\partial \eta}{\partial x}=2\dfrac{\partial}{\partial \xi}+\dfrac{\partial}{\partial \eta}$
$\dfrac{\partial}{\partial y}=\dfrac{\partial}{\partial \xi}\dfrac{\partial \xi}{\partial y}+\dfrac{\partial}{\partial \eta}\dfrac{\partial \eta}{\partial y}=\dfrac{\partial}{\partial \xi}.$
$\begin{eqnarray*}
\dfrac{\partial^2}{\partial x^2}&=&\dfrac{\partial}{\partial x}\left(\dfrac{\partial}{\partial x}\right)\\
&=&\left(2\dfrac{\partial}{\partial \xi}+\dfrac{\partial}{\partial \eta}\right) \left(2\dfrac{\partial}{\partial \xi}+\dfrac{\partial}{\partial \eta}\right)\\
&=&4\dfrac{\partial^2}{\partial \xi^2}+4\dfrac{\partial^2}{\partial \eta\partial \xi}+\dfrac{\partial^2}{\partial \eta^2}.
\end{eqnarray*}
$
$\dfrac{\partial^2}{\partial y^2}=\dfrac{\partial^2}{\partial \xi^2}$
$\dfrac{\partial^2}{\partial y\partial x}=2\dfrac{\partial^2}{\partial \xi^2}+\dfrac{\partial^2}{\partial \xi \partial \eta}$
Thus, we have that:
$\left(\dfrac{\partial^2}{\partial x^2}-4\dfrac{\partial^2}{\partial y\partial x}+4\dfrac{\partial^2}{\partial y^2}\right)u=0\implies u_{\eta \eta}=0.$
This is correct? :)