If $f(\frac{x}{y})=\frac{f(x)}{f(y)} \, , f(y),y \neq 0$ and $f'(1)=2$ then $f(x)=$? If $f(\frac{x}{y})=\frac{f(x)}{f(y)} \, , f(y),y \neq 0$ and $f'(1)=2$ then $f(x)=$?

I am not sure where to begin, any hints on starting and steps is apreciated.

Thank you
$f(x/y)=f(x)/f(y),\forall x,y$


if you take $x=y$ you obtain $f(1)=1$. As a consequence $f(1/y)=1/f(y)$.
$x=0$ and $f$ not constant implies $f(0)=0$.
if $a,b \neq 0$ then $f(ab)=f(a/(1/b))=f(a)/(f(1/b))=f(a)f(b)$. If for an element $x \neq 0$ we have $f(x)=0$ then $f$ would be constant, which contradicts $f'(1)=2$. 
In particular you have $f(x^2)=(f(x))^2&gt;0$.
Define for $x&gt;0$ $g(x)=\log(f(e^x))$. Then $$g(x+y)=\log(f(e^xe^y))=\log(f(e^x))+\log(f(e^y))=g(x)+g(y)$$
This means that $g$ satisfies a Cauchy functional equation and it is continuous (since you assume $f$ to be differentiable at $1$). This means that there exists $a$ such that $g(x)=ax$.
This leads to: $\log(f(e^x))=ax \Rightarrow f(e^x)=(e^{x})^a \Rightarrow f(x)=x^a$ for $x&gt;0$.
Now the condition $f'(1)=2$ implies that $a=2$ and $f(x)=x^2$ for $x&gt;0$.
Because $f(x^2)=f(x)^2$ even for negative $x$ you have $f(-x)=\pm f(x)$.
Linear functions satisfied this function-equation trivially.

Let $f(x)=2x$ then $f'(x)=2$ for all $x$ and $f\left(\frac{x}{y}\right)=\frac{f(x)}{f(y)}$.

That's all :-)

R