How do we get the function $f(y')=f(bx-ay)$? The following is from Partial Differential Equations by Strauss:

Let us solve $au_{x} + bu_{y} = 0$, 
  where $a$ and $b$ are constants not both zero.



Coordinate Method Change variables (or “make a change of coordinates”;
  Figure $2$) to
$$x' = ax + by$$ $$y' = bx − ay.$$ 
Replace all $x$ and $y$ derivatives by $x'$ and $y'$ derivatives. By the chain rule,
$u_{x} = \dfrac{∂u}{∂x} = \dfrac{∂u}{∂x'}\dfrac{∂x'}{∂x}+ \dfrac{∂u}{∂y'}\dfrac{∂y'}{∂x}=au_{x'}+bu_{y'}$
$u_{y} = \dfrac{∂u}{∂y} = \dfrac{∂u}{∂y'}\dfrac{∂y'}{∂y}+ \dfrac{∂u}{∂x'}\dfrac{∂x'}{∂y}=bu_{x'}-au_{y'}$
Hence $au_{x} + bu_{y} = a(au_{x'} + bu_{y'}) + b(bu_{x'} − au_{y'}) = (a^{2} + b^{2})u_{x'}$. So, since $a^{2} + b^{2} \ne 0$, the equation takes the form $u_{x'} = 0$ in the new (primed) variables. Thus the solution is $u = f(y')
= f (bx − ay)$, with $f$ an arbitrary function of one variable. 

Setting $u_{x'}=0$ in both $u_{x}$ and $u_{y}$, I get that $u_{x}=bu_{y'}$ and $u_{y}=-au_{y'}$. Am I supposed to integrate $u_{x}$ and $u_{y}$ with respect to $x$ and $y$, respectively to get that $u=bxu_{y'} + g(y)$ and $u=-ayu_{y'} + h(x)?$ This implies that $g(y)=-ayu_{y'}$ and $h(x)=bxu_{y'}$ and thus $u=bxu_{y'}-ayu_{y'}=(bx-ay)u_{y'}.$ But how do we get that $u=f(bx-ay)?$
 A: You have to solve $u_{x'} = 0$. The PDE $u_{x'}(x', y') = 0$ has the general solution $u(x', y') = f(y')$. Now you just have to make the transformation undone, i. e., replace $x'$ and $y'$. You obtain $u(x, y) = f(bx - ay)$ since $y' = bx - ay$.
A: By taking a linear sum of the $2$ linear equations for $u_x$ and $u_y$, and using the given fact that not both of $a$ and $b$ are $0$ so $a^2 + b^2 \neq 0$, this means the factor of $u_{x'} = 0$. Thus, this shows that $u$ has no component of $x'$. As such, $u = f(y')$ for some function $f$. Since $y' = bx - ay$, this gives that $u = f(bx - ay)$.
As for what you're trying to do, note that you can't get from $u_x = bu_{y'}$ to $u = -ayu_{y'} + g(y)$ and from $u_y = -au_{y'}$ to $u = -aya_{y'} + h(x)$ by integrating wrt to $x$ and $y$ respectively because $u_{y'}$ has components of $x$ and $y$. You can see that by solving the $2$ equations given for $u_{x'}$, for instance, by multiplying the first one by $b$, subtracting the second one multiplied by $a$, then divide by $a^2 + b^2$ to get
$$u_{y'} = \frac{bu_{x} - au_{y}}{a^2 + b^2} \tag{1}\label{eq1}$$
