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 $f(0)=0$.)

Since $f(f(x))=x$, $f$ is bijective and there is a $y_0$ such that $f(y_0)=1$. Now we have
$$f(x+f(x))=y_0f(x)+x$$
$$f(xf(1)+f(x))=f(x)+x$$
which leads to $y_0f(x)+x=xf(1)+f(x)$ for all $x\in\mathbb{Q}$, or
$$(f(1)-1)x=(y_0-1)f(x).$$
Suppose that $y_0\neq 1$, then it turns out that $f(x)$ is a linear function. It is easily verified that in this case the only possibility is $f(x)=x$ which contradicts to $y_0\neq 1$.
Therefore we conclude $y_0=1=f(1)$, and as mastrok noticed we have $f(y+1)=f(y)+1, \forall y$, and equivalently $f(y-1)=f(y)-1,\forall y$.
The next step is for every $x\neq 0$, suppose $f(y)=1/x$ and we get
$$yf(x)+x=f(1+f(x))=f(f(x))+1=x+1,$$
and hence $y=1/f(x)$. That is,
$f(1/x)=1/f(x),\forall x\neq 0.$
Now consider the set $S=\{x\mid f(x)=x\}$. We already know $0\in S$, and $S$ is closed under operations 'plus one', 'minus one' and 'reciprocal'. Now I think it's easy for yourself to check $S=\mathbb{Q}$ by induction with an Euclidean fashion. There's an example here:
$$13/5\rightarrow 3/5 \rightarrow 5/3\rightarrow 2/3\rightarrow 3/2\rightarrow \ldots$$
Update: Since there are ones not familiar with Euclidean algorithm, I try to make it clearer. Use induction on $q$ for fraction $p/q\in\mathbb{Q}$, where $(p,q)=1$ and $q>0$. The Euclidean division turn out that
$$p=aq+b,a\in\mathbb{Z}, 0\leq b<q.$$
And $b/q\in S$ since $q/b\in S$ by inductve hypothesis. Now $p/q=b/q+a\in S$ and the induction is done since the initial case $q=1$ is trivial.
