If $ f(x \cdot f(y) + f(x)) = y \cdot f(x) + x $, then $f(x)=x$ Let $ f : \mathbb{Q} \rightarrow \mathbb{Q} $ be a function which has the following property:
$$ f(x \cdot f(y) + f(x)) = y \cdot f(x) + x \;,\; \forall \; x, y \in \mathbb{Q} $$
Prove that $ f(x) = x, \; \forall \; x, y \in \mathbb{Q} $.
So far, I've found that $f(f(x)) = x$, $f(0) = 0$ and $f(-1) = -1$. 
(For $f(0)=0$, we substitute $x=0$ to arrive at $f(f(0))-yf(0)$ identically $0$ for all rational $y$; for $f(f(x))=x$, we substitute $y=0$ and use $f(0)=0$. For $f(-1) = -1$, substitute $x=y=-1$ to get $f(0)=-f(-1)-1$, and use $f(0)=0$.)

\begin{align}
P(0, y): \; & f(f(0))=yf(0)=0. \\
\Rightarrow \; & f(0)=0. \\
\ \\
P(x, 0): \; & f(f(x))=x. \\
\ \\
f(P(x, y)): \; & f(x)+xf(y)=f(x+yf(x)). \\
f(y) \to y: \; & f(x)+xy=f(x+f(x)f(y)). \tag 1 \label 1 \\
\ \\
P(-1, -1): \; & f(f(-1)-f(-1))=-1-f(-1). \\
\Rightarrow \; & f(-1)=-1. \\
\ \\
x \to -1 \text{ from } \color {#4995CF} {(\ref 1)}: \; & -1-y=f(-1-y). \\
y \to -x-1: \; & f(x)=x.
\end{align}
