$$ x^3u_x-u_y=e^{2u} $$
I think that this is a semilinear and I used the characteristic equation which I get $-1/(x^3)$ and from there I get $c(s)= y-1/(2x^2)$ but I'm not sure if the approach is correct. Can you tell me If this is correct so far or if it is a semilinear equation?
$$x^3u_x-u_y=e^{2u}$$ You are right, a first characteristic equation is $y-\frac{1}{2x^2}=c_1$. The Charpit-Legendre system of characteristic ODEs is : $$\frac{dx}{x^3}=\frac{dy}{-1}=\frac{du}{e^{2u}}$$ $\frac{dx}{x^3}=\frac{dy}{-1}$ leads to the first characteristic equation that you found : $$y-\frac{1}{2x^2}=c_1$$ A second characteristic equation comes from solving $\frac{dy}{-1}=\frac{du}{e^{2u}}$ : $$y-\frac12 e^{-2u}=c_2$$ The general solution expressed on the form of implicit equation $c_2=f(c_1)$ is : $$y-\frac12 e^{-2u}=f(y-\frac{1}{2x^2})$$ $f$ is an arbitrary function.
Or equivalently on explicit form : $$u(x,y)=-\frac12\ln\left|2y+F(y-\frac{1}{2x^2})\right| $$ $F=-2f\:$ is an arbitrary function (to be determined according to some boundary conditions which are missing in the wording of the question).