How can I prove or disprove that there exists a function such that... Suppose we have a function $f$ of $bx-ay$ where $a$ and $b$ are two real constants, if we have for example $e^{bx-ay}$  then obviously it is a function of $bx-ay$.
Can we find a function $f$ such that: $f(bx-ay) = ax-by$? in other words what operations we should operate on $bx-ay$ to get $ax-by$? how can I proceed?
 A: No. Here's why. Consider $a = 1, b = -2$ and look at  $x = 0, y = 2$ and $x = 1, y = 0$. For each of these $bx - ay = -2$. But for the first, $ax - by = 4$, while for the second $ax - by = 1$. 
Since the function $f$ can only take on one value for the argument $-2$ (because of the definition of "function"), it must take on the value either $4$ or $1$, but not both. 
A similar argument works for almost any other pair of values for $a$ and $b$; the only exception I can see is $a = b = 0$, when it's easy to build $f$ (but not interesting: it's the everywhere-zero function). 
A: Here is my solution:
we can put $z = bx-ay $ and $w = ax-by$
So we have: $$f(z) = w$$
from the previous relation we get:
\begin{cases}
x = \dfrac{aw-bz}{a^2-b^2}\\
y = \dfrac{bw-az}{a^2-b^2}
\end{cases}
if $w$ do not depends on $z$ then $\dfrac{\partial w}{\partial z}=0$
hence:
$$\dfrac{\partial w }{\partial z} = \dfrac{\partial w }{\partial x }\dfrac{\partial x}{\partial z }+\dfrac{\partial w}{\partial y}\dfrac{\partial y }{\partial z } = 0$$
we have: $\dfrac{\partial w}{\partial x } = a$, $\dfrac{\partial w}{\partial y } = -b$, $\dfrac{\partial x}{\partial z } = \dfrac{-b }{a^2-b^2 }$ and $\dfrac{\partial y }{\partial z } = \dfrac{-a }{a^2-b^2 }$
Therefore:
$$\dfrac{-ab }{a^2-b^2 }+\dfrac{ab }{a^2-b^2 }=0$$
This is valid only if $a^2\neq b^2 \Rightarrow a = \pm b$
Conclusion $ax-by$ depends on $bx-ay$ only if $a = \pm b$
