Finding $f$ such that $f(xy) = xf(y)+yf(x)-2xy$ given $f'(1)=3$ Let $f$ be a differentiable function satisfying the relation                $$f(xy) = xf(y)+yf(x)-2xy$$ where $x, y&gt;0$ and $f'(1)=3$ then prove that the equation f(x) = k has two solutions in
 $k\in(-e^{-3}, 0)$

I tried differentiating this function but couldn't get anything from it. How to proceed here?
Hint: you can compute $f$ explicitly. Let $g(x)=\frac {f(x)} x -2$ and verify that $g(xy)=g(x)+g(y)$. Do you know how to find all continuous functions satisfying this equation?. [$f(x)=x(c\log\, x+2)$].
Hint: you can compute f explicitly. Let g(x)=f(x)x−2 and verify that g(xy)=g(x)+g(y). Do you know how to find all continuous functions satisfying this equation?. [f(x)=x(clogx+2)].