To determine if a 2 variable symmetric function is addition formula of one variable function or not? Since $$f(x+y)=f(y+x)$$, So an addition formula must be symmetric.
If we define $$f(x+y)=U(f(x),f(y))$$
If we define $f(x)=p$ and $f(y)=q$
$$f(x+y)=U(p,q)$$
and because of $f(x+y)=f(y+x)$, 
$$f(x+y)=U(f(x),f(y))=U(f(y),f(x))$$
Thus $U(p,q)$ is a two variable symmetric function
$$U(p,q)=U(q,p)$$
An example:
$$f(x+y)=f(x)f(y)(f(x)+f(y))$$ 
$U(x,y)=xy(x+y)$ kernel can be a candidate of an addition formula of $f(x)$ because it is a symmetric function.
But If we extend it for 3 components $f(x+y+z)$
$$f(f^{-1}(x)+f^{-1}(y)+f^{-1}(z))=zxy(x+y)(z+xy(x+y))$$ 
The result is asymmetrical, so $$f(x+y)=f(x)f(y)(f(x)+f(y))$$   cannot be an addition formula
Other example is $tan(x)$ addition formula , It has symmetric kernel too
$U(x,y)=\frac{x+y}{1-xy}$
 and after 3 component adding, the result is also symmetric.
$$tan(x+y+z)=\frac{tan x+tan y+tan z -\tan x\tan y\tan z}{1-(\tan x \tan y+\tan x \tan z +\tan y \tan z)}$$
Is there a formula to determine if a symmetric function $U(x,y)$ is kernel of an addition formula of a function without testing as I made above to add 3 components to determine it manually ?
Thanks
 A: The relevant result is a theorem of Weierstrass which says that if $f$ is meromorphic (of one variable), and satisfies an addition theorem
$$f(x+y)=F(f(x),f(y)),$$
then $f$ is elliptic (possibly degenerate). The converse is also true.
The a priori assumption that $f$ is meromorphic can be substantially relaxed with
the same conclusion. Thus we are reduced to describing all possible addition formulas for elliptic curves, and I do not think that there is an explicit answer. 
A: I found out a way to determine if a two variable symmetric function is an addition function of one variable function or not.
Let's assume that we know $$\frac{df(x)}{dx}=G(f(x))$$
We can write that 
$$\int \frac{d(f(x))}{G(f(x))}=\int dx$$
$$\int \frac{d(f(x))}{G(f(x))}=x+c$$
$$\int_{f(0)}^{f(x)} \frac{dz}{G(z)}=x$$
$$\int_{f(0)}^{f(y)} \frac{dz}{G(z)}=y$$
$$\int_{f(0)}^{f(x+y)} \frac{dz}{G(z)}=x+y$$
$$\int_{f(0)}^{f(x+y)} \frac{dz}{G(z)}=\int_{f(0)}^{f(x)} \frac{dz}{G(z)}+\int_{f(0)}^{f(y)} \frac{dz}{G(z)} $$
If we define $$f(x+y)=U(f(x),f(y))$$
$$\int_{f(0)}^{U(f(x),f(y))} \frac{dz}{G(z)}=\int_{f(0)}^{f(x)} \frac{dz}{G(z)}+\int_{f(0)}^{f(y)} \frac{dz}{G(z)} $$
If we define $f(x)=p$ and $f(y)=q$
$$\int_{f(0)}^{U(p,q)} \frac{dz}{G(z)}=\int_{f(0)}^{p} \frac{dz}{G(z)}+\int_{f(0)}^{q} \frac{dz}{G(z)} $$
If we derivate both sides over $p$
 $$\frac{\partial {U(p,q)}}{\partial p} \frac{1}{G(U(p,q))}=\frac{1}{G(p)} $$
 $$\frac{\partial {U(p,q)}}{\partial p} =\frac{G(U(p,q))}{G(p)} $$
If we derivate both sides over $q$
 $$\frac{\partial {U(p,q)}}{\partial q} \frac{1}{G(U(p,q))}=\frac{1}{G(q)} $$
 $$\frac{\partial {U(p,q)}}{\partial q} =\frac{G(U(p,q))}{G(q)} $$
And Finally divide two results that we got 
$$\frac{\frac{\partial {U(p,q)}}{\partial p} }{\frac{\partial {U(p,q)}}{\partial q} }=\frac{G(q)}{G(p)} $$
We can continue to cancel $G(x)$ 
 $$\ln [\frac{\partial {U(p,q)}}{\partial p}] -\ln[ \frac{\partial {U(p,q)}}{\partial q}] =\ln [G(q)]-\ln [G(p)] $$
If we derivative both sides over $p$
 $$\frac{\frac{\partial^2 {U(p,q)}}{\partial p^2}}{\frac{\partial {U(p,q)}}{\partial p}} -\frac{\frac{\partial^2 {U(p,q)}}{\partial p\partial q}}{\frac{\partial {U(p,q)}}{\partial q}} =-\frac{G'(p)}{G(p)} $$
If we derivative both sides over $q$, we get a condition without $G(x)$
$$\frac{\partial}{\partial q}[\frac{\frac{\partial^2 {U(p,q)}}{\partial p^2}}{\frac{\partial {U(p,q)}}{\partial p}} -\frac{\frac{\partial^2 {U(p,q)}}{\partial p\partial q}}{\frac{\partial {U(p,q)}}{\partial q}}] =0 $$

Let's test the example :
$$f(x+y)=f(x)f(y)(f(x)+f(y))$$ 
$$f(x+y)=U(f(x),f(y))=f(x)f(y)(f(x)+f(y))$$ 
If we define $f(x)=p$ and $f(y)=q$
$$f(x+y)=U(p,q)=p q(p+q))$$ 
$$\frac{\partial {U(p,q)}}{\partial p}=q(p+q)+p q=q^2+2 p q$$
$$\frac{\partial {U(p,q)}}{\partial q}=p(p+q)+p q=p^2+2 p q$$
To check condition 
$$\frac{\frac{\partial {U(p,q)}}{\partial p} }{\frac{\partial {U(p,q)}}{\partial q} }=\frac{G(q)}{G(p)} $$
$$\frac{G(q)}{G(p)}=\frac{q^2+2 p q}{p^2+2 p q} $$
It is impossible to find a $G(x)$ that satisfies the relation. 

If we test other example  :
$$f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$$ 
$$f(x+y)=U(f(x),f(y))=\frac{f(x)+f(y)}{1-f(x)f(y)}$$ 
If we define $f(x)=p$ and $f(y)=q$
$$f(x+y)=U(p,q)=\frac{p+q}{1-pq}$$ 
$$\frac{\partial {U(p,q)}}{\partial p}=\frac{1+q^2}{(1-pq)^2}$$
$$\frac{\partial {U(p,q)}}{\partial q}=\frac{1+p^2}{(1-pq)^2}$$
To check condition 
$$\frac{\frac{1+q^2}{(1-pq)^2}}{\frac{1+p^2}{(1-pq)^2}}=\frac{G(q)}{G(p)}$$
$$G(x)=c(1+x^2) $$  where $c$ is constant
Thus we can write that 
$$f'(x)=c(1+f^2(x))$$  
