<s>[INST] Determine all functions satisfying $f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$ 
Let $\Bbb R $ be the set of real numbers .Determine all functions $f:\Bbb R \rightarrow \Bbb R $ satisfying the equation
  $$f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$$ for all real numbers $x$ and $y$.

My work 
$$f(x + f(x + y)) + f(xy) = x + f(x + y) +yf(x)$$
Setting $y=0$
$$f(x + f(x)) + f(0) = x + f(x )$$
$$\implies f(x + f(x))  = x + f(x ) -f(0)$$
$$\implies f(t)  = t -f(0)$$
Setting $x=0$
$$f(f(y)) + f(0) = 0 + f(y) +yf(0)$$
$$f(f(y)) =  f(y) +f(0)(y-1)$$
Setting $y=1$
$$f(x + f(x + 1))  = x + f(x + 1)$$
 [/INST] Here's a clean solution invoking fixed points.
Denote $P(x,y)$ as the assertion that $f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$.
$P(0,0) \implies f(f(0))=0$
$P(x,1) \implies$ $x+f(x+1)$ is a fixed point.
$P(x,0) \implies f(x+f(x))+f(0)=x+f(x)$.
$P(0,x) \implies f(f(x))+f(0)=f(x)+xf(0)$.
Take $P(0,y+f(y+1))$. Using that $y+f(y+1)$ is a fixed point, we get $f(y+1)+y-1=0$ or $f(0)=0$.
Therefore, $f(0)=0$ or $f(x)=2-x$ for all $x$.
If $f(0)=0$, we now have by $P(x,0)$ that $x+f(x)$ are fixed points.
$P(x,-x)$ gives $f(x)+f(-x^2)=x-xf(x)$ and swapping $x$ and $-x$ gives $f(-x)+f(-x^2)=-x+xf(-x)$.
Subtract these two to get $f(x)-f(-x)=2x-xf(x)-xf(-x)$.
Plug $x=1$, get $f(1)=1$. Then $P(1,-1)$ gives $f(-1)=-1$.
Now $P(1,y)$ gives $f(1+f(y+1))+f(y)=1+f(y+1)+y$.
Set $y=x+f(x+1)$. Then $y, y+1$ are all fixed points.
This gives us $f(y+2)=y+2$, so $x+f(x-1)$ is a fixed point.
$P(x,-1)$ gives $f(x+f(x-1))+f(-x)=x+f(x-1)-f(x)$ and using that $x+f(x-1)$ is a fixed point gives $f$ is odd.
Since $f(x)-f(-x)=2x-xf(x)-xf(-x)$ and $f(x)=-f(-x)$, we get $f(x)=x$.
Therefore $f(x) \equiv x$ or $f(x) \equiv 2-x$.
</s>