Cute problem! Don't know why it's P3... Let $P(x,y): 2f(x+g(y))=f(2g(x)+y)+f(x+3g(y)).$ step1) $f(\mathbb{R})$ is a finite set. $P(0,y): 2f(g(y))=f(2g(0)+y)+f(3g(y)) \rightarrow f(2g(0)+y)=f(3g(y))-2f(g(y))$. since $g(y)$ can have finite number of values when $2g(0)+y$ can be any number in $\mathbb{R}$, $f(\mathbb{R})$ is a finite set. Let $f(\mathbb{R})={a_1,a_2, ...,a_n}$ $(a_1<a_2< ... <a_n)$. step2) If $f(a)=a_n$, $f(a+2g(y))=a_n$ for all $y$ in $\mathbb{R}$. Just see $P(a-g(y),y)$. step3) $f(x)=constant.$ if $n\ge 2:$ There exist $b,a$ such that $f(b)=a_{n-1}$ and $f(a)=a_n$. $P(b-3g(a),a): 2f(b-2g(a))=f(2g(b-3g(a))+a)+f(b)=a_{n-1}+a_n$ which is a contradiction!

The answer is constant functions only, which obviously works. Let $P(x,y)$ denote the assertion in the problem, $Im(g)=\{a_1, \cdots,a_n\}$, and let $T_j=\{ y | g(y)=a_j\}$ Claim 1: $|Im(f)| \le |Im(g)|$ $P(x, t_j)$ where $t_j$ varies over all elements of $T_j$ gives $f(2g(x)+t_j)=2f(x+a_j)-f(x+3a_j)$ which is constant as $t_j$ varies over $T_j$. Since $T_1 \cup T_2 \cup \cdots \cup T_n = \mathbb{R}$, it follows that $Im(f) \subseteq \{ f(2g(x)+t_j) | 1\le j\le n\}$ so $Im(f)$. Let $Im(f)=\{ x_1,\cdots,x_m\}$. WLOG $x_1=\max\{ x_1,\cdots,x_m\}, x_m=\max\{x_1,\cdots,x_m\}$ and $X_j=\{ x| f(x)=x_j\}$. Now that $Im(f)$ is finite, if $f(x+g(y))$ is maximal, so are $f(x+3g(y))$ and $f(2g(x)+y)$ Select $x\in X_1-a_j$. Since $x+a_j\in X_1, f(x+a_j)=x_1$. By maximality of $x_1$, it follows that $f(2g(x)+t_j)=x_1$ for all $t_j\in T_j$ (i.e. $2g(x)+T_j \subseteq X_1$) If $x+g(y)\in X_1$ then it must follow that $x+3g(y)\in X_1$ and $2g(x)+y\in X_1$. This means $X_1+2a_j \subseteq X_1$ because if $z\in X_1, P(z-g(y),y)$ gives $f(z+2g(y))=x_1$. (1) Now, select $x\in X_1, z\in X_m$. $P(x-g(y),y): 2g(x-g(y))+y \in X_1$. By (1), $2g(x-g(y))+2g(z-g(y))+y\in X_1$ $P(z-g(y),y): 2g(z-g(y))+y \in X_m$. By (1), $2g(x-g(y))+2g(z-g(y))+y\in X_m$ It follows that $x_1=x_m$, so $f$ is constant.

Clarification: CANBANKAN wrote: Select $x\in X_1-a_j$. Since $x+a_j\in X_1, f(x+a_j)=x_1$. By maximality of $x_1$, it follows that $f(2g(x)+t_j)=x_1$ for all $t_j\in T_j$ (i.e. $2g(x)+T_j \subseteq X_1$) If $x+g(y)\in X_1$ then it must follow that $x+3g(y)\in X_1$ and $2g(x)+y\in X_1$. This means $X_1+2a_j \subseteq X_1$ because if $z\in X_1, P(z-g(y),y)$ gives $f(z+2g(y))=x_1$. (1) Here for a set $X$ and a real $c$, $X+c$ doesn't mean $X\cup \{c\}$ but $\{a+c| a\in X\}$ Similarly, $X-c$ doesn't mean $X\backslash \{c\}$ but $\{a-c| a\in X\}$. Sorry for the possible confusion, and hopefully this clears things up. In the future, when I use notation, I will define them.

We see that f has finitely many values by $P(0,x)$ as also done by above, we define $\alpha, \beta$ such that $f(\alpha)=a, f(\beta)=b$ where $a,b$ are respectively the maximum and minimum values that f can take. $$P(\alpha - g(\beta - g(\alpha)), \beta - g(\alpha)) \implies f( \alpha + 2g(\beta - g(\alpha)))=a $$$$P(\beta - g(\alpha), \alpha) \implies f(\alpha + 2g(\beta - g(\alpha))) = b$$so $\alpha = \beta$, $f(x)=c$ for some real $c$.

Slightly different solution: Claim $f$ has finite range. Proof Notice $2f(x+g(y))-f(x+3g(y))=f(2g(x)+y)$, move $y$ over all reals, the LHS takes finitely many values, and the RHS is $f$ of any real number. As a corrolary of this, $f$ has an absolute maximum. Say $k$ is cool if $f(k)$ achieves this maximum. The idea is that if $x+g(y)$ is cool, then $x+3g(y)$ and $2g(x)+y$ are also cool. Claim For all real $y$, there's an $x$ such that $2g(x)+y$ is cool. Proof Basically choose the $x$ such that $x+g(y)$ is cool. Claim If $k$ is cool, then $k+2g(y)$ is also cool for any $y\in\mathbb{R}$. Proof Plug $x=k-g(y)$ in to the equation. Now just notice that combining the two claims, we get that if $s$ is the sum of the elements of $g(\mathbb{R})$, then $y+2s$ is cool for all $y$, thus all reals are cool and $f$ is constant, which indeed works.

Easy to see that $f$ takes finitely many values. Let $f(X)$ be the maximum value. Setting $x=X-g(x-3g(X))$ and $y=x-3g(X),$ we deduce $f(X+2g(x-3g(X)))=f(X).$ And setting $x=x-3g(X)$ and $y=X,$ we get $f(x)=f(X)$ for all $x.$ Any constant function works.

CANBANKAN wrote: $x_1=\max\{ x_1,\cdots,x_m\}, x_m=\max\{x_1,\cdots,x_m\}$ Typo, one should be min