Take $x=y$ and you find $f=g$. Next, take $y=0$ and you find $ xh(x)=f(x)-f(0)$. Put $ t(x)=f(x)-f(0)$. Then $ (x-y)t(x+y)=(x+y)(t(x)-t(y))$. Now, take $ y=1$ and you find $ t(x+1)=\frac{x+1}{x-1} (t(x)-t(1))$. Substitute here $x+1$ instead of $x$ and obtain $ t(x+2)=\frac{x+2}{x} (\frac{x+1}{X-1}(t(x)-t(1))-t(1))$. But $ t(x+2)=\frac{x+2}{x-2} (t(x)-t(2))$. From the two relations we find that $t$ is of the form $ ax+b$ and since $ t(0)=0$ we have $t(x)=ax$ and then $ f(x)=ax+b$.

Thats not all the possible functions, note for example $ h(x) = ax$, $ f(x) = g(x) = ax^2 + b$ harazi wrote: From the two relations we find that $ t$ is of the form $ ax + b$ In fact from that tow little guys we get $ t(x)=x^2\times (\frac{1}{2} f(2)-f(1))+ x (2 f(1) - \frac{1}{2} f(2) )$ for $ x\not= 0, 1, 2$ Make the substitucion $ a=\frac{1}{2} f(2)-f(1) \wedge b=( 2f(1) - \frac{1}{2} f(2) \Leftrightarrow f(2)=4a+b \wedge f(1)=a+b$ and we get $ t(x)=ax^2+bx$ for all reals, because it is true for $ x=0$ by defenition of $ t(x)$, and $ f(2)=4a+b=a2^2+b2$, $ f(1)=a+b=a1^2+1b$ Recall $ f(0)=c$ and we get $ f(x)=ax^2+bx+c=g(x)$ and $ h(x)=ax+b$ Now check that $ ax^2+bx+c-ay^2-by-c=(x-y) \times (a(x+y) + b)$ occur for $ x, y$ reals.

Let $P(x,y)$ be the assertion and let $g(0)=c,kf(0)=k$.Then $P(x,0) \Rightarrow f(x)-c=xh(x) \Rightarrow f(x)=xh(x)+c$.$P(0,y) \Rightarrow k-g(y)=-yh(y) \Rightarrow g(y)=k+yh(y)$.Substituting these in the original equation we get $xh(x)-yh(y)+f=(x-y)h(x+y)$ where $f=c-k$.$y=0 \Rightarrow f=0 \Rightarrow c=k$.So the equation may be rewritten as $xh(x)-yh(y)=(x-y)h(x+y)$ Now $P(x,-x) \Rightarrow h(x)+h(-x)=2h(0) \forall x \neq 0$.Substituting $y=-y$ we get $xh(x)+yh(-y)=(x+y)h(x-y)$ $\Rightarrow xh(x)+y(2h(0)-h(y))=(x+y)h(x-y)$ $\Rightarrow (x-y)h(x+y)+2h(0)y=(x+y)h(x-y)$ Substituting $y=(x-1)$ in this equation yeilds $h(2x-1)=(2x-1)(h(1)-2h(0))$.In other words $h(x)=dx$ for some $d \in \mathbb{R}$. We may hence conclude that $f(x)=g(x)=dx^2+c$ for some $d \in \mathbb{R}$ These system clearly satisfies the original FE.