Another second order PDE in canonical form

$$u_{\phi\psi}=\frac{1}{\phi-\psi} u_\psi$$ Yes substitute $z=u_{\psi}$: $$(\phi-\psi)z_{\phi}=z$$ $$(\phi-\psi)z_{\phi}-z=0$$ $$\left ( \dfrac {z}{\phi-\psi} \right)'=0$$ $$ \dfrac {z}{\phi-\psi}=C(\psi)$$ Note that $C$ is not a constant but a function of $\psi$: $$ {u}_{\psi}=C(\psi)(\phi-\psi)$$ Integrating by part gives: $$ {u}{(\psi,\phi)}=g(\psi)(\phi-\psi)+\int g(\psi)d\psi+f(\phi)$$

Letting $V=\partial_{\eta} U$ we get $$\partial_{\xi}V=\frac{1}{\xi-\eta}V$$ This is a straightforward application of characteristics to this semilinear PDE. $$\begin{bmatrix} \partial_{t}[ \xi ](t,s)\\ \partial_{t}[ \eta ](t,s)\\ \partial_{t}[ \zeta ](t,s) \end{bmatrix} =\begin{bmatrix} 1\\ 0\\ \zeta\cdot( \xi -\eta )^{-1} \end{bmatrix}$$ So $\xi(t,s)=t+f_1(s)$, $\eta(t,s)=f_2(s)$ and $$\zeta(t,s)=V(\xi(t,s),\eta(t,s))=f_3(s)(f_1(s)-f_2(s)+t)=f_3(s)(\xi(t,s)-\eta(t,s))$$ From $\eta(t,s)=f_2(s)$ we can write $s=f_2^{-1}(\eta)$

Hence $V(\xi,\eta)=f_4(\eta)\cdot(\xi-\eta)$, where $f_4=f_3\circ f_2^{-1}$. This can be easily verified. Then, $$V=\partial_{\eta}U=f_4(\eta)\cdot(\xi-\eta)$$ We can now directly integrate. $$u(x,y)=U(\xi,\eta)=\int f(\eta)\cdot (\xi-\eta)\mathrm{d}\eta+g(\eta)$$ Which can be integrated by parts from @Aryadeva 's answer. Substitute whatever $\xi,\eta$ are in terms of $x,y$ and you're done.