I saw this question and was convinced that the statement is true. However I attempted to solve the problem by derivating the equation with respect to $x$:
$$\dfrac{\partial f(xy)}{\partial x}y = \dfrac{d f(x)}{d x}.$$
Now if I take the derivative w.r.t. $y$:
$$\frac{\partial^2 f(xy)}{\partial x\partial y}xy + \dfrac{\partial f(xy)}{\partial x} = 0.$$
Let $g(x,y) = \dfrac{\partial f(xy)}{\partial x}$, then:
$$\partial_y (g) xy + g = 0$$
I'm stuck here, ny help is very useful for me, thanks!
Assuming $f$ is differentiable, then it follows \begin{align} \frac{\partial}{\partial x} f(xy) = \frac{\partial}{\partial x}(f(x)+f(y)) \ \ \Rightarrow \ \ yf'(xy) = f'(x) \end{align} for all $y$. Set $x=1$, then it follows \begin{align} yf'(y) = f'(1) \ \ \Rightarrow \ \ f'(y) = \frac{f'(1)}{y} \ \ \Rightarrow \ \ f(y) = f'(1)\ln y + C. \end{align} Hence it follows \begin{align} f(y) = f'(1)\ln y + f(1). \end{align} However, we know that \begin{align} f(1\cdot 1) = f(1)+f(1) \ \ \Rightarrow \ \ f(1) = 0 \end{align} which means \begin{align} f(y) = f'(1)\ln y . \end{align}
Edit: I have assumed differentiability to avoid the potential existence of pathological examples such as the ones given by the Cauchy functional equation $f(x+y) = f(x)+f(y)$.
As stated, the theorem is false. Consider $f(x)=\ln(x+1)$. Restate it as $f(x)+f(y)=f(xy)$ for all $x,y>0$ and $f(1)=0$ and $f$ is continuous then $f(x)=k\ln(x)$ for some constant k.
Let's consider that problem instead. $f(x)+f(x)=f(x^2)$, $f(x)+2f(x)=f(x^3)$, etc. By induction you can prove $nf(x)=f(x^n)$ for all integers n. Making the transformation $x\to x^{1/n}$ you get that $f(x^{1/n})=\frac{1}{n}f(x)$ therefore $f(x^{p/q})=\frac{p}{q}f(x)$ for all integers p,q. Then since we required f to be continuous, $f(x^\alpha)=\alpha f(x)$ for all $\alpha\in\mathbb{R}$. WLOG let $f(e)=1$ then $f(e^x)=x$, in other words f is the inverse of $e^x$ so $f(x)=\ln(x)$ then the whole family of solutions is just scaling that by a constant factor.
You don't necessarily know that $f$ is differentiable. The comments/answers in the question you linked to suggest it's much easier to show that if $g(x+y)=g(x)+g(y)$ for all $x,y$ and $g$ is continuous, then $g(x)=ax$ for some $a$.
The function $f\circ \exp$ satisfies \begin{align*} (f\circ \exp)(x+y))&=f(e^{x+y})\\ &=f(e^xe^y)\\ &=f(e^x)+f(e^y)\\ &=(f\circ \exp)(x)+(f\circ \exp)(y). \end{align*} So you know $f\circ \exp = ax$ for some $a$, so $f(x)=f(e^{\ln x})=(f\circ \exp)(\ln x)=a\ln x$.
You must have $b=0$, because for nonzero $b$, $$a\ln(xy)+b=a\ln x + a\ln y +b \neq (a \ln x + b)+(a \ln y + b).$$
I commited a mistake when derivating. Let $u=xy$, then: $$\partial_x f(xy) = \dfrac{df}{du}y = \dfrac{df}{dx}.$$
Now I derivate w.r.t. $y$:
$$\partial_y \left(\dfrac{df}{du}\right)y + \dfrac{df}{du} = 0.$$
This implies
$$\dfrac{d^2f(u)}{du^2} \overbrace{xy}^{u} + \dfrac{df}{du} = 0$$
Solving this differential equation yields the desired solution.
