Let $\rm P(x,y): f(f(f(x)+y)+y)=x+y+f(y)$. $\rm P(x,0): f(f(f(x)))=x+f(0)$.This means that $\rm f$ is $\rm 1-1$ because if $\rm f(x_1)=f(x_2)$ then $\rm f(f(f(x_1)))=f(f(f(x_2)))\Leftrightarrow x_1+f(0)=x_2+f(0)\Leftrightarrow x_1=x_2$. $\rm P(0,0): f(f(f(0)))=f(0)\Leftrightarrow f(f(0))=0\,\,(*)$ $\rm P(x,-x): f(f(f(x)-x)-x)=f(-x)\Leftrightarrow f(f(x)-x)-x=-x\Leftrightarrow$ $\rm \Leftrightarrow f(f(x)-x)=0\overset{(*)}{=}f(f(0))\Leftrightarrow f(x)-x=f(0)\Leftrightarrow \boxed{\rm f(x)=x+f(0)}$. So we have that : $\rm f(f(x)+y)+y+f(0)=x+y+y+f(0)\Leftrightarrow f(x)+y+f(0)=x+y\Leftrightarrow x+f(0)+y+f(0)=x+y\Leftrightarrow f(0)=0$. So $\boxed{\rm f(x)=x}$

Let $P(x,y)$ be the assertion $f(f(f(x)+y)+y)=x+y+f(y)$. $P(x,0)\implies f(f(f(x)))=x+f(0)$ so $f$ is bijective. \begin{align*} P(x,-x)&\implies f(f(f(x)-x)-x)=f(-x) \\ &\implies f(f(x)-x)-x=-x \\ &\implies f(f(x)-x)=0 \\ &\implies f(x)-x=f^{-1}(0) \\ &\implies f(x)=x+f^{-1}(0)\end{align*}Thus, $f(x)=x+c$ for some constant $c$. We see that the only possible value of $c$ is $c=0$, so $\boxed{f(x)=x}$.

I discussed this problem on my YouTube channel. Link: Video Solution

Interesting Problem! $P(0,0)\rightarrow f(f(f(0)))=f(0)$ $P(f(0),0)\rightarrow f(f(f(f(0))))=f(0)+f(0)=2f(0)\rightarrow f(f(0))=2f(0)$ $P(f(f(0)),0)\rightarrow f(f(f(f(f(0)))))=f(0)+f(f(0))=3f(0) \rightarrow f(f(f(0)))=3f(0) \rightarrow f(0)=3f(0)$ $\therefore f(0)=0$ $P(x,-f(x)) \rightarrow f(f(0)-f(x))=x-f(x)+f(f(x))\rightarrow f(x)=x\ \forall x\in \mathbb{R}$ (Plug back we have $LHS=x+y+y=RHS$)