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
$$f'(x)=c(1+f^2(x))$$