Let f: R - 0 → R be a function satisfying f((x)/(y)) = (f ( x ))/(f...

Given: $f\left(\dfrac{x}{y}\right) = \dfrac{f\left(x\right)}{f\left(y\right)}$

And we know that, if $f\left(\dfrac{x}{y}\right) = \dfrac{f\left(x\right)}{f\left(y\right)}$ then $f\left(x\right) = {x}^{n}$.

So, differentiating the function we get,

${f}^{'}\left(x\right) = n{x}^{n - 1}$ and given ${f}^{'}\left(1\right) = 2024$.

Hence, on comparing we get, $n = 2024$

$⇒ x{f}^{'}\left(x\right) = x \times 2024{x}^{2023}$

$⇒ x{f}^{'}\left(x\right) = 2024{x}^{2024}$

$⇒ x{f}^{'}\left(x\right) - 2024{x}^{2024} = 0$

$⇒ x{f}^{'}\left(x\right) - 2024f\left(x\right) = 0$