Suppose you have for example the PDE $3u_x(x,y)-4u_y(x,y)=0$. If you apply the coordinate transformation $\xi=3x-4y$ and $\eta=4x+3y$, you can transform the PDE into $U_\xi=0$, where $u(x,y)=U(\xi(x,y), \eta(x,y))$.
How do we get the idea for defining $\xi$ and $\eta$? I think it comes from the method of characteristics, but I don't see it.
Sometimes I see PDE where the people just define e.g. $\xi=3x-4y$ and $\eta=y$, but here this doesn't work. This is explained e.g. here.
On
Your PDE tells you that the directional derivative of $u$ in direction $(3,-4)$ is zero, which means that $u(x+3s, y-4s)=u(x,y)$ is constant as a function in $s$. Now there are various ways to chose $s$, mainly to connect the point $(x,y)$ with some boundary point. If you set $s=\frac14y$ then $u(x,y)=u(\frac14(4x+3y),0)$ so that $u$ is identified as a function of $η=4x+3y$.
If the direction field is not constant, you will need to extend this approach to the general method of characteristics.
Taking the PDE
$$ a u_x+b u_y = 0 $$
and changing variables according to
$$ r=a_1 x+b_1 y\\ s = a_2 x+ b_2 y $$
we get
$$ (a a_1+b b_1)u_r+(a a_2+b b_2)u_s = 0 $$
If now we choose
$$ a_1 = b,\; b_1 = -a \;\;\; \mbox{or}\;\;\; a_2 = b,\; b_2 = -a $$
we obtain
$$ u_s = 0 \;\;\; \mbox{or}\;\;\; u_r = 0 $$
which is a simpler PDE.