I am abbreviating the amount of steps I'm showing. First off, this is a hyperbolic PDE. So we have $$\eta=\frac{1}{2}x^2-\frac{1}{2}y^2$$ and $$\xi=\frac{1}{2}x^2+\frac{1}{2}y^2$$.
Putting it into canonical form: $$y^2(x^2u_{\xi\xi}+2x^2u_{\xi\eta}+x^2u_{\eta\eta}+u_\xi+u_\eta)-x^2(y^2u_{\xi\xi}-2y^2u_{\xi\eta}+y^2u_{\eta\eta}+u_{\xi}-u_{\eta})=0$$
And simplifying:
$$u_{\xi\eta}=\frac{\eta}{2(\xi^2-\eta^2)}u_\xi-\frac{\xi}{2(\xi^2-\eta^2)}u_\eta$$
Now we can use integrating factors. From this point on, I'm not as confident in my solution:
Let $$ \large \mu(\eta)=e^{\int\frac{\eta}{2(\xi^2-\eta^2)}\,d\eta}$$ Then we have $$(\mu(\eta)u_\xi)_\eta=\mu(\eta)\frac{\eta}{2(\xi^2-\eta^2)}u_\xi$$
This hopefully yields the following:
First integrating both sides w.r.t $\eta$.
$$\mu(\eta)u_\xi=\left(\int \mu(\eta)\frac{\eta}{2(\xi^2-\eta^2)}u_\xi\,d\eta\right)+F(\xi) $$
Dividing both sides by the integrating factor
$$u_\xi=\frac{1}{\mu(\eta)}\left(\int \mu(\eta)\frac{\eta}{2(\xi^2-\eta^2)}u_\xi\,d\eta\right)+\frac{1}{\mu(\eta)}F(\xi)$$
Now integrating both sides w.r.t $\xi$.
$$u=\int\frac{1}{\mu(\eta)}\left(\int \mu(\eta)\frac{\eta}{2(\xi^2-\eta^2)}u_\xi\,d\eta\right)\,d\xi+\int\frac{1}{\mu(\eta)}F(\xi)\,d\xi+G(\eta)$$
