Let $f:R-\left\{0\right\}\to R$ be a function satisfying $f\left(\frac{x}{y}\right)=\frac{f\left(x\right)}{f\left(y\right)}$ for all $x,y,f\left(y\right)\ne 0$. If $f^{\text{'}}\left(1\right)=2024$, then

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

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

So, differentiating the function we get,

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

Hence, on comparing we get, $n=2024$

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

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

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

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