$$ f(f(x)f(y))+f(x+y)=f(xy). \ \ \ (1)$$
Let $f(0)=c$.
Choose $x=y=0$, we can get
$$f(c^2)=0.\ \ \ (2)$$
Choose $y=0$, we can get
$$f(cf(x))+f(x)=c.\ \ \ (3)$$
Choose $y=\frac{x}{x-1}(x\neq 1)$, we can get
$$f\left(f(x)f\left(\frac{x}{x-1}\right)\right)=0(x\neq 1).\ \ \ (4)$$
When $c=0$, from equation $(3)$ we get $f(x)=0.$
When $c\neq 0$, i.e. $f(0)\neq 0$. Form equation $(4)$ we know there esists $x_0\neq0$ such that $f(x_0)=0.$
We claim that :$x_0=1$.
Otherwise, choose $x=x_0$ in equation $(4)$, we get $f(0)=0$ which is a contradiction.
Combining equation $(2)$, we know $c=1$ or $-1$.
If $c=1$, that is to say $f(0)=1,$
choose $y=1$ in equqtion $(1)$, we get $f(x+1)=f(x)-1$.
So $f(n)=1-n$ and $f(x+n)=f(x)-n$ for all $n\in Z$.
By equation $(1)$ we get
$$f(f(x)f(y)+1)+f(x+y+n)=f(xy+n+1).\ \ \ (5)$$
In the following we prove that: $f$ is injective.
If $f(a)=f(b)$, choose integer $n$ such that $(b-n)^2>4(a-n-1)$,
such that there exists $x_0,y_0$ statisfying
$$x_0y_0+n+1=a,x_0+y_0+n=b.$$
From equation$(5)$ we get $f(x_0)f(y_0)+1=1$, so $f(x_0)=0$ or $f(y_0)=0$
,i.e. $x_0=1$ or $y_0=1$, and this implies $a=b$ which is the injectivity of function $f$.
From equation $(3)$, we know $f(f(x))=1-f(x)$.
On the one hand, $f(f(f(x)))=1-f(f(x))=1-(1-f(x))=f(x);$
on the other hand, $f(f(f(x)))=f(1-f(x))$.
Injectivity of $f$ implies $f(x)=1-x.$
If $c=-1$, we can get $f(x)=x-1$ in the same way as above!
In conclusion, all the solutions of the fucntioanl equation are the following:
$$f(x)=0; f(x)=1-x; f(x)=x-1.$$
We have $g: (-\frac{1}{3},\frac{1}{3}) \to \mathbb{R}$ with $g(x+y) = g(x)+g(y)$. Define $G: \mathbb{R} \to \mathbb{R}$ by $G(x) = Ng(\frac{x}{N})$ where $N \in \mathbb{N}$ is large enough to ensure $|\frac{x}{N}| < \frac{1}{3}$. To see that the definition does not depend on $N$, i.e. to show $Ng(\frac{x}{N}) = Mg(\frac{x}{M})$ for any $M$ with $|\frac{x}{M}| < \frac{1}{3}$, it suffices to show both are equal to $NMg(\frac{x}{NM})$, which is clear from additivity. Let's show $G(x+y) = G(x)+G(y)$ for $x,y \in \mathbb{R}$. Fix $x,y \in \mathbb{R}$, and take $N$ large so that $|\frac{x}{N}|,|\frac{y}{N}|,|\frac{x+y}{N}| < \frac{1}{3}$; then $G(x+y) = Ng(\frac{x+y}{N})$ and $G(x)+G(y) = Ng(\frac{x}{N})+Ng(\frac{y}{N})$, so just use additivity of $g$. Finally, it is clear that $G$ extends $g$.
Best Answer
Let $g=-f$ with $f$ a solution to the equation. Then $$g(xy)=-f(xy)=-f(f(x)f(y))-f(x+y)\\ =g(-g(x)(-g(y)))+g(x+y) = g(g(x)g(y)) +g(x+y)$$
and so $g$ also satisfied the functional equation.