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

We are given a function $$f: \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}$$ satisfying $$f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$$ for all $$x, y$$ with $$f(y) \neq 0$$, and $$f'(1) = 2024$$. We need to find the relationship between $$f$$ and $$f'$$.

Determine the form of $$f(x)$$.

Setting $$x = y$$ in the functional equation:

$$ f\left(\frac{x}{x}\right) = \frac{f(x)}{f(x)} \implies f(1) = 1 $$

Setting $$y = x$$ and $$x = 1$$:

$$ f\left(\frac{1}{y}\right) = \frac{f(1)}{f(y)} = \frac{1}{f(y)} $$

The functional equation $$f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$$ is equivalent to $$f(xy) = f(x) \cdot f(y)$$ (a multiplicative function). For differentiable functions satisfying this, the solution is $$f(x) = x^n$$ for some constant $$n$$.

Determine the value of $$n$$ using $$f'(1) = 2024$$.

If $$f(x) = x^n$$, then $$f'(x) = nx^{n-1}$$.

$$ f'(1) = n \cdot 1^{n-1} = n = 2024 $$

Therefore, $$f(x) = x^{2024}$$.

Find the differential equation satisfied by $$f$$.

With $$f(x) = x^{2024}$$:

$$ f'(x) = 2024x^{2023} $$

Multiply both sides by $$x$$:

$$ xf'(x) = 2024x^{2024} = 2024 \cdot f(x) $$

Therefore:

$$ xf'(x) - 2024f(x) = 0 $$

Verification: We can verify by substituting: $$x \cdot 2024x^{2023} - 2024 \cdot x^{2024} = 2024x^{2024} - 2024x^{2024} = 0$$. Confirmed.

The correct answer is Option (1): $$xf'(x) - 2024f(x) = 0$$.