Let $A$ be the set of all functions $f:\mathbb{R}\to\mathbb{R}$ such that $f(xy)=xf(y)$ for all $x,y\in \mathbb{R}$. If $f\in A$ then show that $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb{R}$.
My attempt:
Given $f(xy)=xf(y)$ for all $x,y\in \mathbb{R}$. Since for all $x,y\in \mathbb{R}$, $x+y\in \mathbb{R}$,
so we have $$f((x+y)z)=(x+y)f(z)\quad \text{for all}\; x,y,z\in \mathbb{R}.$$ Now we choose $z=1$, $$f((x+y)1)=(x+y)f(1)\quad \text{for all}\;x,y\in \mathbb{R}.$$ $$\therefore f(x+y)=xf(1)+yf(1)=f(x)+f(y)\quad \text{for all}\;x,y\in \mathbb{R},$$ since $xf(y)=f(xy)$ and put $y=1$ gives $xf(1)=f(x)$.
Is my proof correct? If it is, then I am looking for alternative proof.
Put $f(x+y)=f(x)+f(y)+\phi(x+y)$.
Clearly $f(0)=0$ so you have $f(x)=f(x+0)=f(x)+f(0)+\phi(x+0)$ Thus $$f(x)=f(x)+\phi(x)\space\text {for all } x\iff \phi(x)=0\space\text {for all } x$$ Thus $f(x+y)=f(x)+f(y)$