If $g(x) = x+c$ for some constant $c$, then $f(x+y+c) = x+y+c$ so it is the identity, so ignore this case. Then, there exists some $k$ such that $p = g(k)-k-g(0) \neq 0$. Then we have that $f(g(k) + y) = f(k+y) = f(g(0) + k+y)$ so $f$ is periodic with period $p$. Then $g(y+p) = f(g(0)+y+p) = f(g(0)+y) = g(y)$ so $g$ is also periodic, as desired. $\blacksquare$

Let $P(x,y)$ denote the given assertion. Notice that by $P(x,0)$, $f(g(x)) = g(x)$, so if $g$ is surjective, then $f$ is the identity function. Now we assume $g$ is not surjective. $P(0,x): f(x+g(0)) = g(x)$. Case 1: $g$ is injective. Then if $f(a) = f(b)$, then $P(0,a-g(0))$ compared with $P(0,b-g(0))$ gives $g(a-g(0)) = g(b-g(0))$, so $a=b$. The original FE can be rewritten as $f(g(x) + y)= f(x +y+g(0))$, so using $f$ injective, we get $g(x) = x + g(0)$, which is surjective, contradiction. Case 2: $g$ is not injective. Then $g(a) = g(b)$ for some $a\ne b$. Comparing $P(a,x)$ and $P(b,x)$ gives $g(a+x) = g(b+x)$ for all reals $x$, so $g(x) = g(x + (b-a))$, which implies $g$ is periodic.

Solution: The solution will be a casework whether $g$ is injective or not. We shall prove for all injective $g$, $f$ must be the identity and $g$ must be periodic if $g$ is non-injective. Let $P(x,y)$ be the assertion to the Functional Equation. Case 1: $g$ is an injective function. $P(x,-g(x))$ gives $f(0) = g(x-g(x))$. The final trick is to consider $P(0,-g(0))$ which gives you $f(0) = g(-g(0))$. By injectivity you get $g(x) = x + g(0)$. Putting this back into the original equation, we yield $f(x+y+g(0)) = x+y+g(0)$ which gives $f \equiv \text{id}$ as desired. Case 2: $g$ is a non-injective function. By assumption, there exists distinct $a,b$ in $\mathbb{R}$ such that $g(a) = g(b)$. Consider $P(x,a)$ and $P(x,b)$. This would give \[f(g(a) + y) = f(g(b) + y) = g(a+ y) = g(b+ y)\]This nicely rearranges to $g(y) = g(y+a-b)$ which shows $g$ is periodic function. We have exhausted all cases so the solution is complete. $\blacksquare$

Let $P(x,y)$ be the given assertion. $P(0,x-g(0))\implies f(x)=g(x-g(0))$ $P(x,y-x)\implies g(g(x)-x-g(0)+y)=g(y)$ Now either $g$ is periodic, or $g(x)=x+g(0)$ in which case $f(x+y+g(0))=x+y+g(0)$ and $f$ is the identity.

Observe that $f(g(x))= g(x)$ and $f(g(0)+y)=f(g(y))$. Suppose that for each pair of real numbers $(x,p)$ we have $$g(x) \neq f(g(x)+p) = g(x+p)$$. Thus $f$ must be injective by below, so $g(y)= g(0)+y \implies f(g(0)+y)=f(g(y))= g(y) = g(0) + y$, hence $f$ is the identity function. If there exists such pair, then $$g(x+p) = g(x) \implies g(x)+a = g(x+p)+a \implies f(g(x)+a) = f(g(x+p)+a) \implies g(a+x) = g(x+a+p)$$for any arbitrary real $a$. Hence $g$ is periodic.

Let us assume that g is not periodic. We have to prove that in this case, f is the identity function. Let $P(x, y)$ denote the given assertion. $P(0, y)$ gives us $f(y+g(0)) = g(y) \implies f(y) = g(y-g(0))$. Substituting this in the original equation, $g(y + g(x) - g(0)) = g(y+x)$. Note that if for some $r$, if $g(r) - g(0) \ne r$, then $|g(r) - g(0) - r|$ is a multiple of the period of $g$, so $g$ becomes periodic. Therefore for all r, $g(r) - g(0) = r \implies g(r) = r + g(0)$. Now $f(r) = g(r-g(0)) = r - g(0) + g(0) = r$, which yields that $f$ is indeed the identity function.

Plug in $y=0$ to get $f(g(x))=g(x)$, which gives \[f(g(x)+y)=g(x+y)=f(g(x+y)).\]We will do cases on whether $f$ is injective. If it is, then we have $g(x)+y=g(x+y)$, and plugging in $x=0$ gives us $g(0)+y=g(y)$. Thus, $g(x)=x+f(0)$ for all $x$, which gives \[f(g(x)+y)=f(x+y+f(0))=g(x+y)=x+y+f(0).\]Thus, $f$ must be the identity. If $f$ is not injective, then $f$ clearly cannot be the identity. There must be some $a\neq b$ such that $f(a)=f(b)$. For each $x$, there is some $y_1$ and $y_2$ such that $g(x)+y_1=a$ and $g(x)+y_2=b$, so \[g(x+y_1)=f(a)=f(b)=g(x+y_2),\]giving $g(x)=g(x+(y_2-y_1)) = g(x+(b-a))$ for all $x$. Thus, $g$ is periodic.

Suppose that $g$ is not periodic then let $P(x,y)$ the assertion of the F.E. $P(x,y-g(x))$ gives $f(y)=g(y+x-g(x))$, now let $x$ vary among all reals to get that $g(x)-x$ is constant therefore $g(x)=x+c$ which by replacing means $f(x+y+c)=x+y+c$ or $f= \text{id}$ thus we are done .

Nice and easy FE We will treat two cases: $g$ is injective and $g$ is not injective. Case 1: $g$ is injective plugging in $y=-g(x)$ we get: $f(0)=g(x-g(x))$, so, by the injectivity of $g$, $x-g(x)$ is equal to a constant $c$, $\forall x \in \mathbb{R}$. Therefore $g$ is also surjective, so setting $y=0$ we get $f(g(x))=g(x)$, therefore $f$ is the identity function. Case 2: $g$ is not injective Then, there exist some $a, b \in \mathbb{R}$ such that $a \neq b$ but $f(a)=f(b)$. Therefore, setting $x=a$ and then $x=b$ we get that $g(a+y)=g(b+y)$, $\forall y \in \mathbb{R}$. Therefore, $g$ is periodic with period $(a-b)$, as desired