# $f:\mathbb{R} \rightarrow \mathbb{R}$, $f(xf(y)+f(x))=2f(x)+xy$

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 ($$f(f(x)) = x$$)

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.