Given, $f:\mathbb{R}\rightarrow \mathbb{R}$, which satisfies $f\left(\dfrac{x}{y}\right)=\dfrac{f(x)}{f(y)}$, where f is continuous and differentiable $\forall x,y\in\mathbb{R},f(y)\neq0$, $f'(1)=2$ and $f(x)\neq0$,$\forall x\neq0$. Find all such possible fuctions. We can easily figure it out that $f(0)=0$,
If we put $x=0$, then we get, $f(y)f(0)=f(0)$, and since $f'(1)\neq 0\Rightarrow f(0)=0$.
Thanks in anticipation.
First, plug in $x=y=1$ to see that $f(1) = 1$. Now, since the function is differentiable, we compute it's derivative. By definition, (assume $x\neq 0)$ $$f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim_{h \rightarrow 0} f(x)\cdot \frac{\frac{f(x+h)}{f(x)}-1}{h} = \lim_{h \rightarrow 0} \frac{f(x)}{x}\cdot \frac{{f(1+h/x)}-f(1)}{h/x}$$ As $h\rightarrow 0$, $h/x \rightarrow 0$. Thus we have $$f'(x) = \frac{f(x)}{x}\cdot f'(1) = \frac{2}{x}f(x)$$
Now just solve the differential equation. Multiply $1/x^2$ on both sides to get $$\frac{1}{x^2}f'(x) - \frac{2}{x^3}f(x)=0$$ This is equivalent to $$\frac{d}{dx}\left(\frac{1}{x^2}f(x)\right) = 0$$ Thus we see that $f(x)=Cx^2 (x\neq 0)$, where $C$ is a constant.
Note: I assumed that $x$ is never 0. So I will have to find the correct value for $x=0$ but I don't see how I can calculate it from $f(x/y)=f(x)/f(y)$. Perhaps the domain is not $\mathbb{R}$?
Edit: The question was edited, so that $f(0)=0$. Then $f(x)=Cx^2$ would certainly work.
Edit: Using the fact $f'(1)=2$, we get $c=1$, hence, $f(x)=x^2$.