Denote $P(x,y)$ the assertion that $(y+1)f(x)+f(xf(y)+f(x+y))=y$ holds. Denote $f(0)=a$. $P(0,x)$ implies $f(f(x))=(1-a)x-a$. Now, either $a=1$ or either $f$ is bijective. So, two cases: 1. $f$ is bijective. Then $P(x,-1)$ implies $f(xf(-1)+f(x-1))=-1$ and hence by injectivity $xf(-1)+f(x-1)=c$ for some $c \in \mathbb{R}$. But then clearly $f(x)=mx+b$ and plugging this into the original equation we find that the only possible such function is $f(x)=-x$ which is indeed a solution. 2. $a=f(0)=1$. Then $f(f(x))=-1$ for all $x$. In particular, for $x=0$ and $x=f(k)$ for some $k$ this yields $f(1)=f(-1)=-1$. $P(1,1)$ implies $f(f(2)-1)=3$ and hence $f(3)=-1$ $P(1,2)$ implies $-3+f(f(2)-1)=2$ i.e. $f(f(2)-1)=5 \ne 3$. Absurd! Hence the only solution to the equation is $\boxed{f(x)=-x}$.

If X=0 we find that( y+1)f(0)+f(f(y))=0 if X be constant we find that f(xf(y)+f(x+y))=y-(y+1)f(x) So we find that f contains all the real numbers if X=0 and f(y1)=f(y2). We find that y1=y2. If X=p Y=0 such that f(p)=0we find that f(pf(0))=0 so we find that f(0)=1 Now X=0. So we find that (y+1)+ f(f(y))=y so we are done

Let $P(x,y)$ denote the assertion. $P(x,-1): f(xf(-1)+f(x-1))=-1$ $P(0,y): (y+1)f(0)+f(f(y))=y$. So $f(f(y))=y-(y+1)f(0)=y(1-f(0))-f(0)$ Hence if $f(a)=f(b), f(f(a))=f(f(b))$, so $f(0)=1$ or $a=b$. Hence $f$ is either injective or $f(0)=1$. If $f$ is injective, $xf(-1)+f(x-1)=c$ for a constant $c$ so $f(x-1)=c-xf(-1)$, $f(x)=c-(x+1)f(-1)$. Checking, only $f(x)=-x$ works. Otherwise, $f(0)=1$. Notice that this implies $f(f(y))=-1$ $P(-1,1): 2f(-1)+f(-f(1)+f(0))=1$ Hence -2+f(2)=1, $f(2)=3$. Since $f(f(2))=-1,f(3)=-1$. Now $P(1,2): 3f(1)+f(f(2)+f(3))=2$, or $-3+f(2)=2$, or $0=2$ which is clearly false.

We claim the only answer is $f(x) = -x$ Let $P(x, y)$ denote the assertion. Now we claim that $f$ is injective. For the sake of contradiction assume otherwise. Then $f(a) = f(b)$ with $a \neq b$ exist. From $P(0, a)$ and $P(0, b)$ we have, \begin{align*} (a+1) \cdot f(0) + f(f(a)) = a\\ (b+1) \cdot f(0) + f(f(b)) = b. \end{align*}From this we conclude $f(0) = 1$. However then $P(0, y)$ gives that $f(f(y)) = -1$. Then from $P(x, 0)$ we find that, \begin{align*} -f(x) = f(x+f(x)). \end{align*}Finally using $P(a, 0)$ and $P(b, 0)$ we are able to derive that $a = b = 0$, contradiction. Thus $f$ is injective. Now from $P(x, 0)$ we have, \begin{align*} x = x \cdot f(0) - f(x) \end{align*}which implies $f(x) = x(f(0) - 1) = cx$. Plugging back into the original equation we must have $c = -1$ so we are done.