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So far I've only got that $f(x) = x + 1 \qquad\forall x \in\mathbb{R}$ is probably the only solution, and that

Substitute (1,y): $f(f(y)+f(1))=y+2f(1) \implies f\text{ surjective}$

$f(x)=f(y) \implies f(f(x)+f(1))=f(f(y)+f(1)) \implies x+2f(1)=y+2f(1) \implies f \text{ injective}$

Let $c$ be such that $f(c)=0$.

Substitute (c,0): $f(cf(0)+f(c))=2f(c)+(0)(c)\implies f(cf(0))=f(c) \implies cf(0)=c \implies c=0 \text{ or } f(0)=1$

If $c=0$, $f(0)=0$. Substitute (x,0): $f(xf(0)+f(x))=2f(x)+x(0)\implies f(f(x))=2f(x) \implies f(y)=2y$ (since $\forall y, \exists x$ such that $f(x)=y$). But this fails when substituted back into the original equation. Hence $f(0)=1$

I've looked at the very similar $f(xf(y)−f(x))=2f(x)+xy$, but I still fail to solve this question without f being involutive ($f(f(x)) = x$)

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As you've noted, $f$ is bijective and $f(0)=1$. Substituting $x,y=0$ shows that $f(1)=2$.

Then substituting $x,y=-1$ shows that $f(-1)=0$. Letting $x=-1$ shows $f(-f(y))=-y$ hence $f(-2)=-1$.

Now let $y=-2$, which shows that $$f(f(x)-x)=2f(x)-2x=2(f(x)-x)$$ By surjectivity, there exists a $z$ such that $f(z)=f(x)-x$ which means, by the above, $f(f(z))=2f(z)$. But letting $x=z$ and $y=-1$, we find that $f(f(z))=2f(z)-z$. Hence $2f(z)=2f(z)-z\implies z=0$.

So for all $x\in\mathbb{R}$, $$f(0)=f(x)-x\implies f(x)=x+f(0)=x+1$$ which is easily seen to be the only function satisfying the equation.

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