Solve the functional equation $F(xy)+xy=xy(f(x)+f(y))$,$\forall x,y,\in (0,\infty)$

Question

Let $f:(0,\infty)\to \mathbb{R}$ be a function such that $f(1)=0$. If $f$ has an antiderivative $F: (0,\infty)\to \mathbb{R}$ such that $$F(xy)+xy=xy(f(x)+f(y)),\forall x,y,\in (0,\infty),$$find $f$.
This problem is from the Romanian magazine "Gazeta Matematica", No.6-7-8/2018.
I would like you to review my two solutions to this problem and post any others that you may find.
Solution 1: For $y=1$ in the given relation we get that $$F(x)+x=xf(x),\forall x\in (0,\infty).(*)$$ From here it follows that $f(x)=\frac{F(x)}{x}+1,\forall x\in (0,\infty)$ and therefore $f$ is differentiable.
By differentiating $(*)$ we get that $$f(x)+1=f(x)+xf'(x),\forall x\in (0,\infty)\iff f'(x)=\frac{1}{x},\forall x\in (0,\infty)$$
Hence, $f(x)=\ln x+C$,$\forall x\in (0,\infty)$, where C is a constant. Since $f(1)=0$, it follows that $C=0$ and $f(x)=\ln x,\forall x\in (0,\infty)$.
Solution 2: As in Solution 1, we deduce that $f$ is differentiable, so in particular it is continuous.
From $f(x)=\frac{F(x)}{x}+1,\forall x\in (0,\infty)$, we get that $f(xy)=\frac{F(xy)}{xy}+1,\forall x,y\in (0,\infty)\iff F(xy)+xy=xyf(xy),\forall x,y\in (0,\infty)$.
After substituting this back into the original equation we get that $$xyf(xy)=xy(f(x)+f(y)),\forall x,y\in (0,\infty)\iff f(xy)=f(x)+f(y),\forall x,y\in (0,\infty)$$
Now, this is Cauchy's logarithmic equation and since $f$ is continuous it follows that $f(x)=C\ln x, \forall x\in (0,\infty)$, where $C$ is a constant.
Substituting back into the original equation we get that $C=1$, so $f(x)=\ln x$, $\forall x\in (0,\infty)$