We claim that $f(x)=x$ is the only such function. Throughout, let $P(x,y)$ denote the assertion $f(x+f(x)+xy)=2f(x)+xf(y)$. Firstly, $f(0)=0$. If $f(0)\neq 0$, then there would exist a nonzero constant $d$ such that $f(d)=0$. Then $P(d,0)$ gives $df(0)=0$, which is a contradiction. Furthermore, there is no other value apart from $0$ for which $f(x)=0$. If there existed a nonzero $c$ such that $f(c)=0$, $P(c,-1)$ would imply that $cf(-1)=0$, so $f(-1)=0$. Hence we have from $P(x,-1)$ that $f(f(x))=2f(x)$ for all $x$. Now take a $u$ such that $f(u)=-1$. Then $0=f(-1)=f(f(u))=2f(u)=-2$, a contradiction. This establishes our claim. As obtained earlier, $P(x,-1)$ gives $f(f(x))=2f(x)+xf(-1)$. Substituting $x\to -1$ here tells us that $f(-1)$ is a fixed point. So $f(-1)=f(f(f(-1)))=2f(f(-1))+f(-1)^2=2f(-1)+f(-1)^2\implies f(-1)=-1$ since $f(-1)$ is nonzero. From this, $P(-1,0)$ directly implies $f(-2)=-2$ Then using the fact that $-1$ is a fixed point, we have that $f(f(x))=2f(x)-x$ (from $P(x,0)$) and so $P(f(x),-2)$ gives us \begin{align*} 2f(f(x))+f(x)f(-2) &= 0 \\ 4f(x)-2x-2f(x) &=0 \\ 2f(x) &= 2x \\ f(x) &= x \end{align*}as desired. From this, it is easy to verify that the identity function satisfies both conditions. Hence the only such function is $f(x)=x$ for all real $x$.

The second statement basically implies f is surjective,right?

Let $P(x,y) : f(x+f(x)+xy) = 2f(x)+xf(y)$ $P(x,-1) = f(f(x))=2f(x)+xf(-1) $ . So $f$ is injective . $P(-1,-1) = f(f(-1))=f(-1)$ . So $f(-1)=-1$ . By $(2)$ we can get $\exists \alpha : f(\alpha)=0$ . By $P( \alpha, \alpha)$ we get $f(0)=0$ . $P(-1,-2) = f(-2)=-2$ $P(-1,x) = f(-2-x)=-2-f(x)$ . ( ) $P(-2,x) = f(-4-2x)=-4-2f(x)=2(-2-f(x))=2f(-2-x)$ ( ) By surjective and ( ) and ( ) we get $f(2x)=2f(x)$ . ( ) $P(x,0) = f(x+f(x))=2f(x)$ . ( ) By ( ) and ( ) we can see $f(x)=x$ . $\blacksquare$