These are 2 not-working examples:
Obviously, if $f=g$, then $g(x)=1f(x)$ so there is exists $k\in\mathbb{R}$
$f(x)=0.5x², g(x)=2x²$, then $0.5x²\cdot 4x=2x³=x\cdot2x²$, but $g(x)=4f(x)$, so there exists a $k\in\mathbb{R}$ again
etc. I got this puzzle from a friend of mine, and he told me it is solvable, but he also told me it's not that easy. I think, f and g cannot be polynomials, I don't have a concrete proof, neither can I explain why, but it's just a very strong feeling. It has to be something a little more complicated. Does someone have a tip? I'm gonna test some sine functions now. The function may not be a piecewise-defined function, though.
You have a good friend.
An answer could be like this: $f(x) = \begin{cases} x^2, \forall x \geq 0;\\ -x^2, \forall x \leq 0; \end{cases}$ and $g(x) = x^2$.