Find all function $f:\mathbb Q \rightarrow \mathbb Q $ $$f(xf(y)+f(x))=yf(x)+x , \forall x,y \in \mathbb Q$$
So , for $x = y = 0$ we get that $f(f(0))=0$ . For $x = 0$ $\implies f(f(0))=yf(0) , \forall y \in \mathbb Q \iff yf(0)=0 ,\forall y \in \mathbb Q \implies f(0)=0$. Now if we make $y=0 \implies f(f(x))=x ,\forall x \in \mathbb Q \implies$ f bijective(don't really used it but it will probably come in handy at some point).It seems that the function that checks is $f(x)=x$ and the fact that the function is defined on rational numbers, we will probably have to arrive at a Cauchy functional equation. If you have any idea or a solution, I'm here to listen. Thanks!