I'll keep working

I'll assume $g(0)=0$ (in a previous post I solved otherwise). Fact 1: $f(x)=g(x)$ for every $x$.

Fact 2: Either $z=0$ is the only zero of $f(z)=g(z)=0$, or $f(x)=g(x)=0$ for every $x$.

[I edited it to make it easier.]

After my last two posts, the functional equation that still remains to be solved is $f(x+yf(x))=f(x)+xf(y)$, and the only zero is at $f(0)=g(0)=0$. For every $x \neq 0$, $f(x) \neq 0$ and we can define $h(x)= \frac{-x}{f(x)}$. $y=h(x)$ $\to$ $f(0)=f(x)+xf(h(x))$ $\to$ $f(h(x))= \frac{1}{h(x)}$. It would be helpful to prove that there's only one possible value for $c$ with $f(\frac{1}{c})=c$. In that case, $h(x)=\frac{1}{c}=\frac{-x}{f(x)}$ and the only solutions left would be $f(x)=-cx$. And checking with the functional equation, $-cx+c^2xy=-cx-cxy$ $\to$ $c^2=-c$ $\to$ $c=0$ or $c=-1$. The only new solution would be $g(x)=f(x)=x$. Unfortunately, our assumption is wrong, because with this function there's not only one possible value for $c$ with $f(\frac{1}{c})=c$. There are two, $c=-1$ and $c=1$. So, we're back to the beginning

Let's solve the case when $g(x)=f(x)$ It obvious that $f(x) \equiv 0$ is a solution.Assume that there exists $s \in R$ such that $f(s) \neq 0$. $y=0, x=s \Longrightarrow f(s)=f(s)+sf(0) \Longrightarrow f(0)=0$ Let $f(t)=0 \Longrightarrow y=s, x=t \Longrightarrow f(t+sf(t))=f(t)+tf(s) \Longrightarrow t=0$.So $f(t)=0$ iff $t=0$ $x=y=-1 \Longrightarrow f(-1-f(-1))=0 \Longrightarrow f(-1)=-1$ $y=1 \Longrightarrow f(x+f(x))=f(x)+xf(1)$ $(1)$ $y=-1,x=1 \Longrightarrow f(1-f(1))=f(1)-1$ $1-f(1)+f(1-f(1))=0 \Longrightarrow f(1-f(1)+f(1-f(1)))=0 \Longrightarrow$ from $(1)$ $\Longrightarrow 0=f(1-f(1)+f(1-f(1)))=f(1-f(1))+(1-f(1))f(1) \Longrightarrow f(1)=1$ $x=1 \Longrightarrow f(1+y)=1+f(y)$ $y \Longrightarrow \frac{1}{f(x)} \Longrightarrow f(x+1)=f(x)+xf(\frac{1}{f(x)})$ $\Longrightarrow f(x)+1=f(x)+xf(\frac{1}{f(x)}) \Longrightarrow f(\frac{1}{f(x)})=\frac{1}{x}$ $y=x, x \Longrightarrow \frac{1}{f(x)} \Longrightarrow f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}.x$ $f(\frac{1}{f(x)})+1=\frac{1}{x}+\frac{1}{f(x)}.x \Longrightarrow 1=\frac{1}{f(x)}.x$ So $f(x)=x$.

cefer wrote: $y=x, x \Longrightarrow \frac{1}{f(x)}\Longrightarrow f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}.x$ If I have understood your substitutions correctly,you should got $f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}*f(x)$ in place of $f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}.x$,shouldn't you?

$f(x+yf(x))=f(x)+xf(y)$ I solved this here. It's quite complex.

cefer wrote: Let's solve the case when $g(x)=f(x)$ It obvious that $f(x) \equiv 0$ is a solution.Assume that there exists $s \in R$ such that $f(s) \neq 0$. $y=0, x=s \Longrightarrow f(s)=f(s)+sf(0) \Longrightarrow f(0)=0$ Let $f(t)=0 \Longrightarrow y=s, x=t \Longrightarrow f(t+sf(t))=f(t)+tf(s) \Longrightarrow t=0$.So $f(t)=0$ iff $t=0$ $x=y=-1 \Longrightarrow f(-1-f(-1))=0 \Longrightarrow f(-1)=-1$ $y=1 \Longrightarrow f(x+f(x))=f(x)+xf(1)$ $(1)$ $y=-1,x=1 \Longrightarrow f(1-f(1))=f(1)-1$ $1-f(1)+f(1-f(1))=0 \Longrightarrow f(1-f(1)+f(1-f(1)))=0 \Longrightarrow$ from $(1)$ $\Longrightarrow 0=f(1-f(1)+f(1-f(1)))=f(1-f(1))+(1-f(1))f(1) \Longrightarrow f(1)=1$ $x=1 \Longrightarrow f(1+y)=1+f(y)$ $y \Longrightarrow \frac{1}{f(x)}\Longrightarrow f(x+1)=f(x)+xf(\frac{1}{f(x)})$ $\Longrightarrow f(x)+1=f(x)+xf(\frac{1}{f(x)}) \Longrightarrow f(\frac{1}{f(x)})=\frac{1}{x}$ $y=x, x \Longrightarrow \frac{1}{f(x)}\Longrightarrow f(\frac{1}{f(x)}+1)=\frac{1}{x}+\frac{1}{f(x)}.x$ $f(\frac{1}{f(x)})+1=\frac{1}{x}+\frac{1}{f(x)}.x \Longrightarrow 1=\frac{1}{f(x)}.x$ So $f(x)=x$. Oops!I have a bad mistake, but here is a solution (i think here isn't any mistake ) From above we got $f(\frac{1}{f(x)})=\frac{1}{x}\Longrightarrow f(x)=1$ iff $x=1$ Replacing $y$ with $\frac{y}{f(x)}$ we get $f(x+y)=f(x)+xf(\frac{y}{f(x)})$ $(1)$ $y \Longrightarrow y+1$ $f(x+y+1)=f(x)+xf(\frac{y+1}{f(x)})$ $f(x+y)+1=f(x)+xf(\frac{y+1}{f(x)})$ $(2)$ From $(1)$ and $(2)$ $1+xf(\frac{y}{f(x)})=xf(\frac{y+1}{f(x)})$ $1+xf(\frac{y}{f(x)})=xf(\frac{y}{f(x)}+\frac{1}{f(x)})$ $(*)$ Applying $(1)$ $1+xf(\frac{y}{f(x)})=x(f(\frac{y}{f(x)})+\frac{y}{f(x)}f(\frac{\frac{1}{f(x)}}{f(\frac{y}{f(x)})}))$ $\frac{f(x)}{xy}=f(\frac{1}{f(x)f(\frac{y}{f(x)})})$ $(**)$ Replacing $x$ with $\frac{1}{f(x)}$ and using the fact that $f(\frac{1}{f(x)})=\frac{1}{x}$ $\frac{f(x)}{xy}=f(\frac{x}{f(xy)})$ Replacing $y$ with $\frac{f(x)}{x}$ we get $1=f(\frac{x}{f(f(x))}) \Longrightarrow f(f(x))=x$ In $(**)$ taking $y=\frac{f(x)}{x}$ we get $1=f(\frac{1}{f(x)f(\frac{1}{x})}) \Longrightarrow f(x)f(\frac{1}{x})=1$ $x \Longrightarrow f(x)$ in $(*)$ $1+f(x)f(\frac{y}{x})=f(x)f(\frac{y}{x}+\frac{1}{x})$ $\frac{1}{f(x)}+f(\frac{y}{x})=f(\frac{y}{x}+\frac{1}{x})$ $f(\frac{1}{x})+f(\frac{y}{x})=f(\frac{y}{x}+\frac{1}{x})$ $f(x)+f(yx)=f(x+yx) \Longrightarrow f(a+b)=f(a)+f(b)$ Using this and main functional equation $f(x)+f(yf(x))=f(x)+xf(y) \Longrightarrow f(yf(x))=xf(y) \Longrightarrow f(xy)=f(x)f(y)$ We got that $f$ is additive and multiplicative so $f(x)=x$

See also here.