$u \in \mathbb{R}$ and $n \in \mathbb{N}$.
To prove that $f(1)=0$, am I on the right track by letting $x=\frac{1}{y}$ to get $f(xy)=f(\frac{1}{y}y)=\frac{1}{y}f(y)+yf(\frac{1}{y})$? If yes, how do I get $0$ from that?
To prove $f(u^n)=nu^{n-1}f(u)$.
There was a hint that said to use $f(u)$ to get $f(u^2)$ and up.
From that I got
$f(u)=f(u)$
$f(u^2)=f(u \cdot u)=2uf(u)$
$f(u^3)=f(u^2 \cdot u)=u^2f(u)+u(2uf(u))=3u^2f(u)$
$f(u^4)=f(u^3 \cdot u)=u^3f(u)+u(3u^2f(u))=4u^3f(u)$
Continuing this to n, I see that it is how you would calculate a derivative.
To figure out $n$, I'm letting $x=u^{n-1}$ and $y=u$. After that I get $f(u^n)=u^{n-1}f(u)+uf(u^{n-1})$. After looking at $u,u^2,u^3$, and $u^4$, I see that $uf(u^{n-1})=u(n-1)u^{n-2}f(u)$. How do I draw that conclusion in the proof and how does that become $nu^{n-1}f(u)$?
Edit: I know that $f(u)=f(u)$ only because I assumed that $f(1)=0$, but I don't how how to get that $f(1)=0$.