$e$ is Euler number.

IMOStarter wrote: Given a real number $\alpha$ and consider function $\varphi(x)=x^2+e^{\alpha x}$ for $x\in\mathbb R$. Find all function $f:\mathbb R\to\mathbb R$ that satisfy: $$f(\varphi(x)+f(y))=y+\varphi(f(x))$$forall $x,y\in\mathbb R$ $\varphi(x)=x^2e^{\alpha x}$

Note that $\varphi(x)$ is a continuous function, $\varphi(x) \ge 0$ for all real $x$, $\varphi(0) = 0$, $\lim_{x \to +\infty} \varphi(x) = +\infty$ or $0$ depending on the sign of $\alpha$, and $\lim_{x \to -\infty} \varphi(x) = +\infty$ or $0$ depending on the sign of $\alpha$. Thus, the range of $\varphi(x)$ is $[0, +\infty)$. Denote the assertion $P(x, y): f(\varphi(x)+f(y))=y+\varphi(f(x))$. $P(0, x-\varphi(f(0))): f(f(x-\varphi(f(0)))) = x$ Thus, $f$ is surjective. Comparing $P(x, c)$ and $P(x, d)$ for distinct reals $c, d$ gives us that $f$ is injective. Let $a = f(0)$ and $b \in \mathbb{R}$ such that $f(b) = 0$. $P(b, 0): f(\varphi(b) + a) = 0 \iff \varphi(b) + a = b$ $P(0, b): a = b + \varphi(a)$ Subtracting the two, we get $\varphi(a) + \varphi(b) = 0$, implying $a = b = 0$. Then $f(f(x)) = x$ and $P(x, 0): f(\varphi(x)) = \varphi(f(x)) \ge 0 \implies f(x) \ge 0$ for all $x \ge 0$. $P(x, f(y)): f(\varphi(x) + y) = f(y) + f(\varphi(x))$ which implies that $f(x + y) = f(x) + f(y)$ for all $x \ge 0$ and $y \in \mathbb{R}$. Then $y \to -x$ gives us $0 = f(x) + f(-x)$ for all $x \ge 0$ (and all $x \in \mathbb{R}$), thus $f$ is odd and $f(x + y) = f(x) + f(y)$ for all real numbers $x, y$. Since $f$ is bounded from below on some interval, by the Cauchy functional equation, $f(x) = cx$ for some $c \in \mathbb{R}$. We get that $\boxed{f(x) = x}$ is the only solution.

IMOStarter wrote: Given a real number $\alpha$ and consider function $\varphi(x)=x^2e^{\alpha x}$ for $x\in\mathbb R$. Find all function $f:\mathbb R\to\mathbb R$ that satisfy: $$f(\varphi(x)+f(y))=y+\varphi(f(x))$$forall $x,y\in\mathbb R$ Kind of scary but nothing u cant deal with. Let $P(x,y)$ the assertion, we claim that $f(x)=x$ is the only function that works. $P(0,x)$ $$f(f(x))=x+\varphi(f(0)) \implies f \; \text{bijective}$$Now let $c$ such that $f(c)=0$ and use $P(0,c)$ $$f(0)=c+\varphi(f(0))$$Now do $P(c,0)$ $$c=\varphi(c)+f(0) \implies \varphi(c)=-\varphi(f(0)) \implies f(0)=0$$$P(0,x)$ $$f(f(x))=x \implies f \; \text{involution}$$$P(x,0)$ $$f(\varphi(x))=\varphi(f(x))$$Now since $x^2$ and $e^{\alpha x}$ are continuos so the function $\varphi(x)$ is continuos and by Intermediate Value Theorem we have that $\varphi$ covers its range which is $[0,\infty]$ Now by $P(x,f(-\varphi(x)))$ $$f(-\varphi(x))=-f(\varphi(x)) \implies f \; \text{odd}$$And now by $P(x,f(y-\varphi(x)))$ $$f(y-\varphi(x))=f(y)+f(-\varphi(x))$$And now by $P(x,f(y))$ $$f(\varphi(x)+y)=f(\varphi(x))+f(y)$$And combining all of these results we get $$f(x+y)=f(x)+f(y) \implies f(x)=ax$$And becuase $f$ is involutive we get $f(x)=x$. Hence $\boxed{f(x)=x}$ is the only function that works, thus we are done

Surprised to see kinda weird function $\varphi (x)=x^2e^{\alpha x}$. Answer: $f(x)=x$ for all real $x$. Proof: Let $P(x,y)$ denote the given assertion, we can easily see that $f$ is bijective. Let $k\in \mathbb{R}$ such that $f(k)=0$, then \[P(0,0)\implies f(f(0))=\varphi(f(0))\]\[P(0,k)\implies f(0)=k+f(f(0)) \implies k=f(0)-f((0))\]\[P(k,0)\implies f(\varphi(k)+f(0))=0\implies k=\varphi(k)+f(0)\]so $\varphi(k)=k-f(0)=-f(f(0))=-\varphi(f(0))$ but this means $0=\varphi(k)+\varphi(f(0))=k^2e^{\alpha k}+f(0)^2e^{\alpha f(0)} \implies f(0)=k=0$ since $\varphi (x)$ is non-negative and $0$ only when $x=0$. Therefore, \[P(0,x)\implies f(f(x))=x\]\[P(x,0)\implies f(\varphi(x))=\varphi(f(x))\]so since $\varphi(x)$ is continuous, non-negative (contains $0$) and it approaches $+\infty$ for large $x$ when $\alpha$ is $\ge 0$ and for negative $x$ when $\alpha$ is $<0$, $\varphi (x)$ is surjective over $[0,+\infty)$. Hence, \[P(x,f(-\varphi(x))\implies f(-\varphi(x))=-\varphi(f(x))=-f(\varphi(x))\]gives $f$ being an odd function. So, let $u=\varphi(x)\ge 0$ and $v\in \mathbb{R},$ \[P(x,f(v))\implies f(u+v)=f(v)+\varphi(f(x))=f(v)+f(\varphi(x))=f(u)+f(v)\]and also \[f(-u+v)=-f(u-v)=-f(u)-f(-v)=f(-u)+f(v)\]give $f(x+y)=f(x)+f(y)$ for all reals $x,y$. Because $f(u)=f(\varphi(x))=\varphi(f(x))\ge 0$ for all non-negative $u$, $f$ takes on positive values for all positive $x$ and thus bounded below for all positive real $x$, so $f$ is linear, plugging $f$ back to our original FE. We get $f(x)=x$ for all real $x$. $\ \blacksquare$