I have a solution to this one (not mine) and I can post it if you want (it's not short but not very long). If you are still interested in trying it yourself here are 2 hints: 1)Prove that $g$ is surjective (that's not very easy). 2)Therefore, it takes value $0$, using it solve the rest of a problem (that's the easier part).

Hello ! My way to prove that g(x) may be not surjective is very short : $(f(x)=0, g(x)=0)$ is a solution in which g is not surjective. -- Patrick

I think that you're totally misunderstanding e.lopes and TomciO! When $f(0) = 0$ (that's the first thing I tried) it's easy to prove that $g$ takes the value $0$. In the other case, I think it may be that $g$ is surjective.

Exactly, in fact, I wasn't clear in my previous post. We need only that $g$ takes value $0$, if $f(0)=0$ then we can do it by hand, in other case we prove that $g$ is surjective.

I found one more nice and short way to solve our equation, my friends! We will prove first that $g(i) = 0$, for any i. If f(0) = 0, we find that $f(x+g(0)) = g(x)$. Take $x =-g(0). g(-g(0)) = f(0) = 0$. If f(0) isn't, is more difficult. See: Take x = 0 in the original equation. We see that f is surjective and g is injective!, because $f(g(y)) = g(0)-y.f(0)$ In original equation, put $x = g(x)$. We find $f(g(x)+g(y)) = g(x)f(y)-yf(g(x)+g(g(x))=g(x)f(y)-y(g(0)-xf(0))+g(g(x))$ This is symetric, so, $g(x)f(y)-y(g(0)-xf(0))+g(g(x)) = g(y)f(x)-x(g(0)-yf(0))+g(g(y))$ (*). Cause f is surjetive, we can take $y = m$, where $f(m) = 0$. Do this in (*)! We find $-f(0)m+g( g(x) ) = g(m) f(x)-f(0)x+g(g(m))$. Putting $g(m) = p$ and $g(g(m))+f(0)m = q$, this becomes $g(g(x)) = pf(x)-f(0)x+q$. Substituting back into (*) we find $g(x) f(y)+pf(x) = g(y)f(x)+pf(y)$ (**) Let y = 0 in (**). We find g(x) = $\frac{(f(0)-p)f(x)}{g(0)}+k$. So, g is surjective! and we can take g(i) = 0! for one real i. _ Now, we will prove that f and g are linears! So, in the original equation, let y = i! We find $g(x) = (i+1)f(x)-f(i)x$. Substituing back into original equality, we find: $f(x+g(y)) = (i+1-y)f(x)+x(f(y)-f(i))$ (***) Let $y = i+1$ in (***). We find that $f(x+r) = x(f(i+1)-f(i))$, where $f = g(i+1)$! So, $f$ is linear! In the equality g(x) = (i+1)f(x)-f(i)x, we see the that g is also linear! _ Now, is easy, let f(x) = ax+b and g(x) = cx+d, and good luck with the computations! The solutions are f(x) = g(x) = 0 and $f(x) = \frac{k(x-k)}{k+1}$ , $g(x) ={k(x-k)}$! for a real k!

suppose that $P(x,y)$ be the following assertion:$f(x+g(y))=xf(y)-yf(x)+g(x)$ $P(x,0):f(x+g(0))=xf(0)+g(x)$. (put $g(0)=c$) $(1)$ $P(x+t+c,y):f(x+c+t+g(y))=(x+t+c)f(y)-yf(x+c+t)+g(x+c+t)$ $P(x+c,y):f(x+c+g(y))=(x+c)f(y)-yf(x+c)+g(x+c)$.now subtracting two equations and use of $(1)$ we have:$tf(0)+g(x+t+g(y))-g(x+g(y))=tf(y)-y(tf(0)+g(x+c+t)-g(x+c))+g(x+t)-g(x)$. so if the previous assertion be $Q(x,y)$,then $Q(x,0):g(x+t+c)-g(x+c)=g(x+t)-g(x)$ $(2)$ then with use of $(2)$ we have $Q(x,1):g(x+t+g(1))-g(x+g(1))=t(f(1)-f(0))$ so for all $x$ we have:$g(x+t)-g(x)=ta$ (put $a=f(1)-f(0)$) put $x=0$ we have:$g(t)=at+c$ so:($put b=f(0)$) $f(x+c)=bx+ax+c$ so $f(x)=(a+b)x+c(1-a-b)$ now by putting it into main equation we conclude that:$a(a+b)=-c(1-a-b)=-b$ and $f(x)=(a+b)x+b,g(x)=ax+c$.or for some real $a\neq -1$.$f(x)=\frac{a(x-a)}{a+1},g(x)=a(x-a)$.

perfect_radio wrote: Find all pairs of functions $ f, g : \mathbb R \to \mathbb R$ such that $ f \left( x + g(y) \right) = x \cdot f(y) - y \cdot f(x) + g(x)\quad (1)$ for all real $ x, y$. Let $f(0)=a$ and $g(0)=b.$ Taking $x=0$ in $(1),$ we get \[f\big(g(y)\big)=-ay+b,\quad \forall y \in \mathbb R. \quad (2)\] There are two cases to consider: Case 1: $\mathbf{a=0.}$ Taking $y=0$ in $(1),$ we get \[f(x+b)=g(x),\quad \forall x \in \mathbb R. \quad (3)\] From this, we deduce that $g({-b})=0.$ Now, replacing $y=-b$ in $(1),$ we have \[f(x)=x\cdot f({-b})+b\cdot f(x)+g(x),\quad \forall x \in \mathbb R. \quad (4)\] Taking $x=-b$ in $(4),$ we get $f({-b})=g({-b})=0$ and hence, the equality $(4)$ can be rewritten as \[(1-b)\cdot f(x)=g(x),\quad \forall x \in \mathbb R. \quad (5)\] Taking $x=0$ in $(5),$ we get $b=g(0)=0.$ Thus, the identities $(2)$ and $(5)$ can be written as \[f\big(g(x)\big)=0,\quad \forall x \in \mathbb R\quad (6)\] and \[f(x)=g(x),\quad \forall x \in \mathbb R. \quad (7)\] From $(6)$ and $(7),$ we get $g\big(g(x)\big)=0,\, \forall x \in \mathbb R.$ Now, replacing $y$ by $g(x)$ in $(1),$ we can easily deduce that $f(x)=g(x)=0,\, \forall x \in \mathbb R.$ These functions satisfy our condition. Case 2: $\mathbf{a \ne 0.}$ From $(2),$ we can easily see that $f$ is surjective and $g$ is injective. Replacing $x$ by $g(x)$ in $(1)$ and using $(2),$ we get \[f\big(g(x)+g(y)\big)=g(x)\cdot f(y)+axy-by+g\big(g(x)\big),\quad \forall x ,\,y \in \mathbb R. \quad (8)\] Changing the position of $x$ and $y$ in $(8),$ we get \[g(x)\cdot f(y)-by+g\big(g(x)\big)=g(y)\cdot f(x)-bx+g\big(g(y)\big),\quad \forall x,\,y \in \mathbb R. \quad (9)\] Replacing $y=0$ in $(1),$ we have \[f(x+b)=ax+g(x),\quad \forall x \in \mathbb R. \quad (10)\] Replacing $x$ by $g(x)$ in $(10),$ we get \[g\big(g(x)\big)+a\cdot g(x)=f\big(b+g(x)\big)=b\cdot f(x)-x\cdot f(b)+g(b),\] or \[g\big(g(x)\big)=b\cdot f(x)-a\cdot g(x)-x\cdot f(b)+g(b),\quad \forall x \in \mathbb R. \quad (11)\] Plugging this result into $(9),$ we get \[\begin{aligned} g(x)\cdot f(y)-by+{}&b\cdot f(x)-a\cdot g(x)-x\cdot f(b)=\\ &=g(y)\cdot f(x)-bx+b\cdot f(y)-a\cdot g(y)-y\cdot f(b).\quad (12)\end{aligned}\] Since $f(b)=b$ (we can easily obtain this by setting $y=0$ in $(2)$), the above identity can be written as \[g(x)\cdot \big[ f(y)-a\big] +b\cdot f(x)=g(y)\cdot \big[ f(x)-a\big] +b\cdot f(y),\] or \[\big[ g(x)-b\big] \big[ f(y)-a\big]=\big[ g(y)-b\big] \big[ f(x)-a\big],\quad \forall x,\,y \in \mathbb R. \quad (13)\] Since $f$ is surjective, there exists $y_0$ such that $f(y_0) \ne a.$ Taking $y=y_0$ in $(13),$ we get \[g(x)-b=\frac{g(y_0)-b}{f(y_0)-a}\big[f(x)-a\big],\quad \forall x \in \mathbb R. \quad (14)\] If $g(y_0)=b,$ then we have $g(x)=b,\,\forall x \in \mathbb R$ which is a contradiction since $g$ is injective. So we must have $g(y_0)\ne b.$ From this and the previous results, we can easily prove that $f$ and $g$ are bijectives. Thus, the exists a unique number $c$ such that $g(c)=0.$ Taking $x=y=c$ in $(1),$ we have $f(c)=g(c)=0.$ Continuously, replacing $y=c$ in $(1),$ we get \[(1+c)\cdot f(x)=g(x),\quad \forall x \in \mathbb R. \quad (15)\] Since $g$ is bijective, it is clearly that $c \ne -1,$ from which it follows that \[f(x)=\frac{1}{1+c}\cdot g(x),\quad \forall x \in \mathbb R. \quad (16)\] Now, replacing $y=1+c$ in $(1)$ and using $(15),$ we get \[f\big(x+g(1+c)\big)=x\cdot f(1+c)=\frac{x\cdot g(1+c)}{1+c},\quad \forall x \in \mathbb R. \quad (17)\] Replacing $x=c-g(1+c)$ in $(17),$ we get \[g(1+c)\cdot \big[c-g(1+c)\big]=0.\] Since $g$ is bijective, we have $g(1+c)\ne 0$ and hence, it follows that $g(1+c)=c.$ From this, we have \[f(x+c)=\frac{cx}{1+c},\quad \forall x \in \mathbb R.\quad (18)\] Replacing $x$ by $x-c$ in $(18),$ we get \[f(x)=\frac{c}{1+c}(x-c),\quad \forall x \in \mathbb R.\] It follows that \[g(x)=c(x-c),\quad \forall x \in \mathbb R.\] These functions satisfy our condition.

Notice f(g(x))=g(0)-f(0)x. (1) If f(0)=0 we get g(-g(0))=0. Otherwise f is surjective and g is injective. taking x=g(x) and considering y as a consant, we get an equation. if we change the value of y in this equation, we find that g(x)=0 for some x. So g(a)=0 for some a always. Then in our original equation f(x)=(f(x+g(a))=xf(a)-af(x)+g(x) so f(x)=(g(x)+xf(a))/(a+1), unless a=-1 but in this case we finish easily. And we get f(a)=af(a)/(a+1). so if a=0 we finish easily, otherwise f(a)=0 and we get g=cf for a constant c. also f is bijective and from here we finish easily. answer: linear f and g=cf for a constant c. additional equations must be satisfied by the coefficients of f, we obtain these relations by just plugging in.

I proved the $f(0) = 0$ the same way as CanVQ. I have a slightly different approach in proving $f(0) \neq 0$, specifically in proving bijectivity of $f$ and $g$ f(0) is not equal to 0 Then $f(g(y)) = -y*f(0) + g(0)$ (1). Lemma: f and g are both bijective Proof: Clearly $f \circ g$ is a bijective function as it's linear. Thus we can immediately conclude from it's injectivity that $g$ is injective and from it's surjectivity that $f$ is surjective. For any $c \in \mathbb{R}$ note that if $f(c) = a$ we have $f(c) = a = \frac{a-g(0)}{f(0)} * f(0) + g(0) = f(g(\frac{g(0)-a}{f(0)}))$ and since $f(0)$ is non-zero we have that there necessarily exists some real number a satisfying $g(a) = c$ for all real c proving surjectivity Now, we had $a,b \in \mathbb{R}$ such that $f(a) = f(b) = c$. Then, we know there exist real m and n such that $g(m) = a$ and $g(n) = b$ Then we would have $f(g(m)) = f(a) = -m*f(0) + g(0)$ and $f(g(n)) = f(b) = -n*f(0)+g(0)$ Thus we have $m=n$ and $a=b$ so f is injective. This completes the lemma. $\square$ Now this means there exists a unique value $a$ such that $g(a) = 0$. Plug in $x=y=a$ into the original equation to get $f(a) = g(a) = 0$. (2) Now, plug in just $y=a$ and use the result from (2) to obtain $f(x) = -a*f(x) + g(x)$ or $(a+1)f(x) = g(x)$ (3) Plugging (3) into (1) reveals $f((a+1)f(x)) = -x*f(0)+(a+1)f(0)$. Now notice that when $x=a+1$ we have that $f((a+1)f(a+1)) = 0$. Thus $f(a+1) = \frac{a}{a+1}$ and thus also $g(a+1) = a$ (4) Now, plug (3) back into the original equation and we get $f(x+(a+1)f(y)) = xf(y) - yf(x)+(a+1)f(x)$. Plug in $y=a+1$ and use (4) and we get: $f(x+a) = \frac{xa}{a+1}$ or $f(x) = \frac{(x-a)a}{a+1}$ and thus from (3) $g(x) = (x-a)a$ for some real constant $a$ which completes the problem. $\square$ (note that since $(a+1)*g(0) = f(0)$ and we have that $g(0) \neq 0$ and $f(0) \neq 0$ this means that $a + 1 \neq 0$ and $a \neq -1$.)

This took way too long T^T Let $x=0$ to get $f(g(y))=-yf(0)+g(0)$ which is linear, so $f(x)$ is surjective and $g(x)$ is injective. Let $y=0$ then $f(x+g(0))=xf(0)+g(x).$ Suppose $x=-g(0)$ then $f(0)=-f(0)g(0)+g(-g(0))$ which implies that $g(-g(0))=f(0)(1+g(0)).$ Thus, if $f(0)=0$ then $g(-g(0))=0.$ Let $x=g(a)$ while $y=b.$ Then \[f(g(a)+g(b))=g(a)f(b)-bf(g(a))+g(g(a))=g(b)f(a)-af(g(b))+g(g(b)).\]Note that we can let $f(a)=0$ to give us $g(g(b))=pf(b)+\text{linear}.$ Thus, $pf(a)-g(a)f(b)=pf(b)-g(b)f(a)$ which can be arranged to form $g(a)=xf(a)+y$ for some constants $x,y$ with $x\neq 0.$ Thus $g$ is surjective unless $g$ is $0$. Now we let $g(y)=0$ so $f(x)(1+y)=xf(y)+g(x).$ Now, let $y=y+1$ to get $f(x+g(y+1))=xf(y+1)-(y+1)f(x)+g(x)=xf(y+1)-xf(y)$ so $f(x)$ is clearly linear, and so $g(x)$ is also linear. Thus, we finish by blindly plugging in coefficients into the original equation to find that $f(x)=\frac{c}{1+c}x-\frac{c^2}{1+c}$ and $g(x)=cx-c^2$ for all $c\neq -1$

Does anyone know the official solution for this? I was like wow...takes forever

Solved with Jeffrey Chen and Max Lu. The answer is \(f(x)\equiv\frac c{c+1}(x-c)\) and \(g(x)\equiv c(x-c)\) for \(c\ne-1\), which work. Now we show these are the only solutions. Let \(P(x,y)\) denote the assertion. Note by \(P(0,x)\) that \[f(g(x))=-f(0)\cdot x+g(0)\quad\forall x.\tag{\(\star\)}\] Claim: We have \[\big[g(y)-g(0)\big]\cdot\big[f(x)-f(0)\big] =\big[g(x)-g(0)\big]\cdot\big[f(y)-f(0)\big].\] Proof. By \(P(g(x),y)\) and \((\star)\), we have \begin{align*} f(g(x)+g(y)) &=g(x)f(y)=f(0)xy-g(0)y+g(g(x)). \end{align*}Combining with \(P(g(y),x)\) gives \[g(y)f(x)+g(g(y))-g(0)x =g(x)+f(y)+g(g(x))-g(0)y.\tag{\(\dag\)}\] Note \((\star)\) gives \(f(g(0))=0\), so \(P(g(x),0)\) and \(P(g(0),x)\) give \[g(g(x))+g(x)f(0)=f(g(x)+g(0))=g(0)f(x)-g(0)x+g(g(0)).\]Substituting \(g(g(x))\) into \((\dag)\), we have \[g(y)f(x)+g(0)f(y)-f(0)g(y)=g(x)f(y)+g(0)f(x)-f(0)g(x),\]which arranges into the desired. \(\blacksquare\) First case: \(f(0)=0\). Then \((\star)\) gives \(f(g(x))=g(0)\) for all \(x\), but \(P(x,0)\) gives \(f(x+g(0))=g(x)\). Hence \[f(f(x))=f(g(x-g(0)))=g(0)\]for all \(x\). In particular \(x=0\) gives \(g(0)=0\), so \(f(x)\equiv g(x)\) and \(f(f(x))\equiv0\). Now \(P(x,f(x))\) gives \(f(x)\equiv0\). Second case: \(f(0)\ne0\). By \((\star)\), \(f\circ g\) is a nonconstant linear function, so \(g\) is injective and \(f\) is surjective. But the claim gives that \(g\) and \(f\) are linear in each other, so \(f\) and \(g\) are both bijective. In particular, \(g(c)=0\) for some \(c\). By \(P(c,c)\), we also have \(f(c)=0\). By \(P(x,c)\), we have \((c+1)f(x)\equiv g(x)\), and by \(P(c,y)\), we have \[f(c+(c+1)f(y))=cf(y).\]Since \(f\) is bijective, we are done.

mathverse06 wrote: Does anyone know the official solution for this? I was like wow...takes forever Here's the kalva solution: https://prase.cz/kalva/short/soln/sh00a3.html. Let's try to solve it myself, too.

Claim 1: If $g(x)=0$ for some $x$ then $f$ and $g$ are linear. Proof. Let $g(z)=0,$ for some real $z.$ $P(x,z)\implies g(x)=(z+1)f(x)-xf(z).$ $P(x,z+1)\implies f\text{ and }g \text{ are linear.}$ Now it's easy to get that, $f(x)=\frac{kx-k^2}{1+k}$ and $g(x)=kx-k^2$ are indeed solutions for all $x\in \mathbb{R}$ and $k\neq -1.$ Claim 2: $\exists z\in \mathbb{R}: g(z)=0.$ Proof. We do casework on $f(0).$ Case 1: $f(0)=0.$ Then $P(x,0)\implies f(x+g(0))=g(x).$ Take $u=-g(0).$ So $g(u)=f(0)=0.$ Q.E.D. Case 2: $f(0)\neq 0.$ Then it is easy to find that $f$ is surjective and $g$ is injective. Comparing $P(g(y),x)$ and $P(g(x),y)$ yields $g(x)f(y)-y(g(0)-xf(0))+g(g(x))=g(y)f(x)-x(g(0)-yf(0))+g(g(y)). ~~~~~(*)$ $\exists v\in \mathbb{R}:f(v)=0.$ $y\to v$ in $(*)$ yields, $g(g(x))=f(x)g(v)-xg(0)+g(g(v))+vg(0).$ $(*)\implies g(x)f(y)+g(v)f(x)=g(y)f(x)+g(v)f(y).$ $y\to 0 \implies g(v)+\frac{g(0)-g(v)}{f(0)}f(x)=g(x).$ We have $g(0)\neq g(v)$ since $g$ is injective and $v\neq 0,$ thus $g$ is surjective. Q.E.D.

review now, will come back and review again later First let $P(x,y)$ be the assertion. From $P(0,y)$ and $P(x,0)$ we get \[f(g(y))=-yf(0)+g(0)\]\[f(x+g(0))=xf(0)+g(x).\] If $f(0)=0$ then $f(g(y))=g(0)$ and $f(f(x))=g(0)$. Hence $x\to 0$ implies $g(0)=0$ and $f(f(x))=0$. In particular $f(x)=g(x)$. Now $P(x,f(y))$ yields \[f(x)=-f(x)f(y)+f(x)\implies f(x)f(y)=0\]hence at most one value of $f$ is nonzero. If $f(a)\neq 0$ then $P(a,a)$ yields \[0=f(a+f(a))=f(a)\]which is a contradiction. This yields the solution $f\equiv g\equiv 0$. If $f(0)\neq 0$ then we find $f$ surjective and $g$ injective. Now the solution will be structured a little differently. If $g(c)=0$ exists then $P(c,c)$ yields $f(c)=0$. Then $P(x,c)$ yields $f(x)=-cf(x)+g(x)$ or $g(x)=(c+1)f(x)$. Hence it makes sense to try and prove $g(c)=0$ exists, or simply that $g$ is surjective. (This also makes sense from the second equation in the first section; there is no reason subtracting $xf(0)$ makes $g(x)$ not surjective.) Notice $P(g(x),y)$ yields \[g(x)f(y)-yf(g(x))+g(g(x))=f(g(x)+g(y))\]and the RHS is symmetric. Now we evaluate the LHS. Notice \[f(g(x))=-xf(0)+g(0)\]\[g(g(x))=f(g(x)+g(0))-g(x)f(0)\]and from $P(g(0),x)$ we get \[f(g(x)+g(0))=g(0)f(x)-xg(0)+g(g(0)).\]Hence the whole expression is equal to \[g(x)f(y)+xyf(0)-yg(0)+g(0)f(x)-xg(0)+g(g(0))-g(x)f(0).\]As the RHS was symmetric switching $x$ and $y$ has no effect. Thus \[g(x)f(y)+g(0)f(x)-g(x)f(0)=g(y)f(x)+g(0)f(y)-g(y)f(0)\]hence \[g(x)f(y)-g(0)f(y)-g(x)f(0)=g(y)f(x)-g(0)f(x)-g(y)f(0)\]or \[[g(x)-g(0)]\cdot [f(y)-f(0)]=[g(y)-g(0)]\cdot [f(x)-f(0)]\]which implies \[[f(x)-f(0)]\div [g(x)-g(0)]=C.\] It follows now that $g$ is surjective and furthermore that $f$ is injective. Noting now that $g(x)=(c+1)f(x)$, we find $P(c,y)$ yields \[f(c+(c+1)f(y))=cf(y)\implies f(c+(c+1)x)=cx.\]Thus we obtain the final solution set: \[f(x)=\frac{c}{c+1}x-\frac{c^2}{c+1}\]\[g(x)=cx-c^2.\]

Assume $f, g$ to be non-constant. Plugging $x=0$ gives us: $f(g(y)) = -f(0)y+g(0) (-)$. Note that $f$ is surjective function. Notice that: $$f(g(x)+g(y)) = g(x)f(y)-yf(g(x))+g(g(x))$$Since LHS is symmetric in $x, y$, swapping them: $$g(x)f(y)-yf(g(x))+g(g(x)) = g(y)f(x)-xf(g(y)) + g(g(y)) (*).$$Consider $t \in \mathbb R$, such that $f(t)=0$. Plugging $t=0$ above with $(-)$, we get: $$g(g(x)) = uf(x)-ax+b$$where $u=g(t), a=g(0), b=g(g(t))+g(0)t$. Plugging back the above equation into $(*)$ along with $(-)$, we get: $$g(x)f(y)+g(t)f(x) = g(y)f(x)+g(t)f(y).$$Thus, we have: $g(x) = \frac{(g(y)-g(t))f(x)}{g(y)}+g(t)=kx+u$. If $k=0$, then $f, g$ are constant. Thus: $k \neq 0$. Notice that: $$g(g(x)) = kf(g(x))+u = k(-f(0)x+g(0))+u = -kf(0)x+kg(0)+u = uf(x)-ax+b.$$If $u \neq 0$, then $f$ is linear and so is $g$. If $u=0$, thus: $g(t)=0$. Therefore: $g(x)=kf(x)$. Plugging $y=t$ in the given equation, we get: $f(x)=xf(t)-tf(x)+g(x) = (k-t)f(x)$. Thus: $k=t+1$. Plugging $x \to kf(x)$ in the given equation, we get: $$f(k(f(x)+f(y))) = kf(x)f(y)-yf(k(x))+kf(x).$$Swapping $x, y$, we have: $$xf(kf(y)) + kf(x) = yf(kf(x))+kf(y).$$Plugging $y=t$ above, we have: $$xf(0)+kf(x) = tf(kf(x)) = (k-1)[ - f(0)x+g(0)].$$Thus, comparing $x$ co-efficient: $k=0$ which leads to contradiction. Thus, we have $f, g$ to be linear. Plugging $f(x)=px+q, g(x)=rx+s$ into the initial equation, we have: $$\boxed{f(x)=\frac{a}{a+1}x-\frac{a^2}{a+1}, g(x)=ax-a^2.}$$

$$f(x+g(y))=xf(y)-yf(x)+g(x)\quad(1)$$We claim that $f(x)=ax+b, g(x)=cx+d$ for all $x\in \mathbb{R}$, where $a,b,c,d$ are some fixed real numbers. First, if $f$ is constant, then let $f(x)=k$ for all $x$. $(1)$ becomes $k=kx-ky+g(x)$. Setting $y=0$ and $y=1$ forces $k=0$. This implies that $g(x)=0$ for all $x$ as well, so the claim holds in this case. From now on, assume that $f$ is not constant. $x\mapsto g(x)$ in $(1)$: $$f(g(x)+g(y))=g(x)f(y)-yf(g(x))+g(g(x))\quad(2)$$$x\mapsto 0$ in $(1)$: $$f(g(y))=-f(0)y+g(0)\quad(3)$$Combining $(2)$ and $(3)$ yields: $$f(g(x)+g(y))=g(x)f(y)+f(0)xy-g(0)y+g(g(x))\quad(4)$$The left-hand side of $(4)$ is symmetric in $x, y$, so we have: $$g(x)f(y)-g(0)y+g(g(x))=g(y)f(x)-g(0)x+g(g(y))\quad(5)$$$y\mapsto 0$ in $(5)$: $$g(g(x))=g(0)f(x)-f(0)g(x)-g(0)x+g(g(0))\quad(6)$$Combining $(5)$ and $(6)$ yields: $$g(x)f(y)+g(0)f(x)-f(0)g(x)=g(y)f(x)+g(0)f(y)-f(0)g(y),$$$$(g(x)-g(0))(f(y)-f(0))=(g(y)-g(0))(f(x)-f(0))\quad(7)$$As $f$ isn't constant, there is some $y_0$ such that $f(y_0)\neq f(0)$. Let $A=(g(y_0)-g(0))/(f(y_0)-f(0))$. $y\mapsto y_0$ in $(7)$ yields: $$g(x)-g(0)=A(f(x)-f(0)),$$$$g(x)=Af(x)-Af(0)+g(0)\quad(8)$$Substituting $(8)$ in $(1)$ gives: $$f(x+g(y))=xf(y)-yf(x)+Af(x)-Af(0)+g(0)\quad(9)$$$x\mapsto x-g(A), y\mapsto A$ in $(9)$: $$f(x)=(x-g(A))f(A)-Af(0)+g(0)$$Therefore, there exist $a,b\in\mathbb{R}$ such that $f(x)=ax+b$ for all $x\in\mathbb{R}$. In light of $(8)$, there also exist $c, d\in\mathbb{R}$ such that $g(x)=cx+d$ for all $x\in\mathbb{R}$. This proves the claim. Substituting $f(x)=ax+b, g(x)=cx+d$ in $(1)$ gives: $$(a-b-c)x+(ac+b)y+(ad+b-d)=0$$This implies that $a-b-c=0,ac+b=0,ad+b-d=0$. Solving in terms of $a$, we have $a\neq 1,b=a^2/(a-1),c=-a/(a-1),d=-a^2/(a-1)^2$. A routine computation shows that all $f, g$ of the form $$f(x)=ax+\frac{a^2}{a-1},\quad g(x)=-\frac{a}{a-1}x-\frac{a^2}{(a-1)^2},$$where $a\in\mathbb{R}, a\neq 1$ really do satisfy $(1)$, so these are the only solutions