Solutions to $(x+y)f(x+y)=xf(x)+yf(y)$

Question

If $x$ is a differentiable function of $t$ and if we define

$$ f(x)=\frac{x^\prime}{x} $$

then $f$ satisfies logarithmic-like properties

1. $f(xy)=f(x)+f(y)$
2. $f(x/y)=f(x)-f(y)$
3. $f(x^n)=nf(x)$

but $f$ also satisfies the non-logarithmic-like property

4. $(x+y)f(x+y)=xf(x)+yf(y)$

Are there any algebraic functions satisfying the functional equation $(4)$?

Note: It is fairly easy to show that $f(x)=f(-x)$.

Answer 1

If $g(x) = x f(x)$, the equation (4) says $g(x+y) = g(x) + g(y)$, i.e. $g$ is additive. See Cauchy's functional equation.

Answer 2

$(x+y)f(x+y)=xf(x)+yf(y) $ (4)

Let $y=x$ you get that $f(2x)=f(x) $

Let $y=2x$ you get $f(3x) = f(x) $

Let $y=nx$ you get $f(nx) = f(x) $

And we know that $f(x) = f(-x) $

$f(1)=f(2)=f(4)=f(5)....$

so f is constant function ?