Find all function $f,g: \mathbb{Q} \to \mathbb{Q}$ such that
\[\begin{array}{l}
 f\left( {g\left( x \right) - g\left( y \right)} \right) = f\left( {g\left( x \right)} \right) - y \\ 
 g\left( {f\left( x \right) - f\left( y \right)} \right) = g\left( {f\left( x \right)} \right) - y \\ 
 \end{array}\]
for all $x,y \in \mathbb{Q}$.
1. **Prove that \( f \) and \( g \) are bijective:**

   Fix \( x \) and consider the first equation:
   \[
   f(g(x) - g(y)) = f(g(x)) - y
   \]
   For \( f \) to be well-defined for all \( y \), \( f(g(x)) - y \) must cover all rational numbers as \( y \) varies. This implies that \( f \) is surjective. Similarly, by fixing \( y \) and varying \( x \), we can show that \( f \) is injective. Therefore, \( f \) is bijective.

   Similarly, for the second equation:
   \[
   g(f(x) - f(y)) = g(f(x)) - y
   \]
   By the same reasoning, \( g \) must be bijective.

2. **Determine the value of \( g(0) \):**

   Let \( y = 0 \) in the first equation:
   \[
   f(g(x) - g(0)) = f(g(x)) - 0 \implies f(g(x) - g(0)) = f(g(x))
   \]
   Since \( f \) is injective, this implies:
   \[
   g(x) - g(0) = g(x) \implies g(0) = 0
   \]

3. **Determine the form of \( f \):**

   Let \( x = 0 \) in the first equation:
   \[
   f(g(0) - g(y)) = f(g(0)) - y \implies f(-g(y)) = -y
   \]
   Since \( g(0) = 0 \), we have:
   \[
   f(-g(y)) = -y
   \]

4. **Prove that \( f \) satisfies the Cauchy functional equation:**

   From the first equation, substituting \( y \) with \( 0 \):
   \[
   f(g(x) - g(y)) = f(g(x)) - y
   \]
   Using the result \( f(-g(y)) = -y \), we get:
   \[
   f(g(x) - g(y)) = f(g(x)) + f(-g(y))
   \]
   This is the Cauchy functional equation, which implies that \( f \) is a linear function over the rationals. Therefore, there exists a constant \( c \) such that:
   \[
   f(x) = cx
   \]

5. **Determine the form of \( g \):**

   Using the second equation:
   \[
   g(f(x) - f(y)) = g(f(x)) - y
   \]
   Substituting \( f(x) = cx \) and \( f(y) = cy \):
   \[
   g(cx - cy) = g(cx) - y \implies g(c(x - y)) = g(cx) - y
   \]
   Let \( z = x - y \):
   \[
   g(cz) = g(cx) - y
   \]
   Since \( g \) is bijective, we can write:
   \[
   g(cz) = \frac{z}{c}
   \]
   Therefore, \( g(x) = \frac{x}{c} \).

The final answer is \( \boxed{ f(x) = cx } \) and \( g(x) = \frac{x}{c} \).