Find the genereal solution to the PDE
$$f_{xx}+2f_{xy}+f_{yy}-2(f_x+f_y)=0,$$
where $f(x,y)\in C^2$ by introducing the new variables $u=e^{x+y}, \ v=e^{x-y}.$ Hint: Note that $u_x=u_y, \ v_x=v, \ v_y=-v.$
I have that
$$ \begin{align} f_{x} &=f_uu_x+f_vv_x=f_uu+f_vv\tag1\\ f_{y}&=f_uu_y+f_vv_y=f_vu-f_vv\tag2\\ \end{align} $$
Differentiating $(1)$ w.r.t $x$ again gives:
$$f_{xx}=u(f_{uu}\cdot u+f_{vv}\cdot v)+f_u\cdot u+\color{red}{v(f_{vu}\cdot u+f_{vv}\cdot v)}+f_v\cdot v= \tag3$$
$$=u^2f_{uu}+2uvf_{uv}+v^2f_{vv}+uf_u+vf_v.$$
Question: Why does $f_{vu}$ come before $f_{vv}$ in $(3)?$ Following the previous logic, should it not be $v(f_{vv}\cdot u + f_{vu} \cdot v)?$
Since $\partial_x=u_x\partial_u+v_x\partial_v=u\partial_u+v\partial_v$ and similarly $\partial_y=u\partial_u-v\partial_v$, $\partial_x+\partial_y=2u\partial_u$. We have $$0=(\partial_x+\partial_y)(\partial_x+\partial_y-2)f=4u\partial_u (u\partial_u f-f)=4u^2\partial^2_u f.$$ Thus $f$ solves our original PDE iff $f$ is linear in $u$.