This is my problem "Find $f:\mathbb{R}\to \mathbb{R}$ s.t. $f(xf(y)-f(x))=2f(x)+xy$ ". I'm not so good at functional equation so all I did until now is subtituting $x,y$ by $1,0$ and I couldn't even calculate a value.
One other approach is that I find 1 solution $f=-x+1$. So I let $g=f+x-1$ and try to prove $g=0$, but subtituting that into the equation make it a mess. Help me please.
This is a really tough problem, I like it very much! I tried to make it clear where I used what, but nevertheless its a bit messy. Feel free to ask if something is unclear!
Let $f$ be a solution of the equation and let $P(x,y)$ e the assertion $f(xf(y)-f(x))=2f(x)+xy$ Then $$ P(1,y):\qquad f(f(y)-f(1))=2f(1)+y\implies f\text{ surjective} $$ Thus let $a,b\in\mathbb{R}$ be such that $f(a)=0$ and $f(b)=1$. $$ P(a,b):\qquad 0=f(af(b)-f(a))=ab $$ Therefore either $a=0$ or $b=0$.
Case 1 : $a=0$ $$ P(1,1):\qquad 0=2f(1)+1\implies f(1)=-\frac12 $$ Thus if we reconsider $P(1,y)$ we obtain $f(f(y)+1/2)=y-1$. Hence $$ P(1,f(y)+1/2):\qquad f(y-1/2)=f(f(f(y)+1/2)+1/2)=f(y)+1/2-1=f(y)-1/2 $$ and with $y=f(x)+1/2$ in the above equation: $$ f(f(x))=f(f(x)+1/2)-1/2=x-3/2 $$ But thus $0=f(f(0))=-3/2$, contradiction.
Case 2 : $b=0$
Then again $P(1,1)$ gives $f(0)=1$. Therefore if we reconsider $P(1,y)$ we get $f(f(y))=y\tag{1}\label{1}$ and thus $f$ is bijective. Now $$ P(x,0):\qquad f(x-f(x))=2f(x)\tag{2}\label{2} $$ $$ P(f(x),0):\qquad f(f(x)-x)=2x\\\stackrel{f()}\implies f(x)-x\stackrel{\eqref{1}}=f(f(f(x)-x))=f(2x)\tag{3}\label{3} $$ $$ P(f(x),1):\qquad f(-x)\stackrel{\eqref{1}}=f(-f(f(x)))\stackrel{P(f(x),1)}=2f(f(x))+f(x)\stackrel{\eqref{1}}=2x+f(x)\tag{4}\label{4} $$ Now $f(-1)\stackrel{\eqref{4}}=2+f(1)=2$ and thus $f(2)=f(f(-1))\stackrel{\eqref{1}}=-1$. Finally we take $$ P(x,2):f(-x-f(x))=2f(x)+2x $$ But $f(-x-f(x))\stackrel{\eqref{4}}=2(x+f(x))+f(x+f(x))$ and thus the above equation simplifies to $f(x+f(x))=0$. But $f$ is injective and we know already that $f(1)=0$, so $$ x+f(x)=1\implies f(x)=1-x\qquad\forall x\in\mathbb{R} $$ Substituting this back into the original equation we see that this is indeed a solution.
Fact 1: $f$ has a root. To see this, take $x \not = 0$ and $y = \frac{-2f(x)}{x}$.
Fact 2: If $f(c) = 0$, then $c = \pm\sqrt{f(0)}$. To see this, set $x=y=c$.
Fact 3: $f(0) = \pm 1$. To see this, let $c$ be a root and set $x=c, y=0$. Then, $f(cf(0)) = 0 \implies cf(0)=\pm\sqrt{f(0)}$ but $c = \pm\sqrt{f(0)}$.
Fact 4: $f(0) = 1$. If $f(0) = -1$, then $x=y=0$ gives $f(1) = -2$. Then $x=1,y=0$ gives $f(-1-(-2))=2(-2) = -4$, a contradiction.
Fact 5: $f(-1) = 2$ and $f(1) = 0$. The first equality follows from $x=0,y=0$. The second then comes from Fact 1, Fact 2, and $f(-1)\not = 0$.
Fact 6: $f(f(x)) = x$ for all $x \in \mathbb{R}$. This follows from setting $x=1$.
Fact 7: $f(-a) = 2a+f(a)$ for all $a \in \mathbb{R}$. This follows from $x=f(a),y=1$ and noting Fact 6 implies $f$ is surjective.
Fact 8: $f(-x-f(x)) = 2x+2f(x)$. This follows from $y=2$.
Fact 9: $f(-x-f(x)) = 2x+2f(x)+f(x+f(x))$. This follows from letting $a = x+f(x)$ in Fact 7.
Fact 10: $f(x) = -x+1$. This follows from Facts 8,9 , Fact 6 showing $f$ is injective, and Fact 5, which gives $f(1) = 0$.
Let's investigate first the affine solutions $x\mapsto ax+b$. The equation rewrites as
$$a(axy+bx-ax-b)+b=2ax+2b+xy\iff (a^2-1)xy+a(b-a-2)x-b(a+1)=0$$
Identifying the coefficients of the terms of the two sides of this polynomial equations we get that $x\mapsto -x+1$ is the only affine solution to the problem.
More painful but with the same method we prove that the only polynomial (power series solution) is $x\mapsto -x+1$
That's all I have for the moment