The equation transforms to \[\frac{f(n+1)}{f(n)} = n + \frac{f(n)}{f(n-1)}\] and induct? But the original equation seems to break when $n = 1$ hmm

Amir Hossein wrote: Let $\mathbb Z^{\geq 0}$ be the set of all non-negative integers. Consider a function $f:\mathbb Z^{\geq 0} \to \mathbb Z^{\geq 0}$ such that $f(0)=0$ and $f(1)=1$, and that for any integer $n \geq 1$, we have \[f(n + 1)f(n - 1) = nf(n)f(n - 1) + (f(n))^2.\]Determine the value of $f(2023)/f(2022)$. $n \mapsto 1 \implies 0 = 1$

I am sorry, I was trying so hard not to make a mistake in translating the questions that I misread $f(0)=1$ for $f(0)=0$. I've fixed in the first post.

First you can prove inductively that $f(n) > 0$ . Then you can write the equation in the form of Thapakazi above. Define $g(n) = \frac{f(n)}{f(n-1)}$. $g(n+1) = g(n) + n$. $g(1) = 1$. Using standard techniques for solving linear recurrence relations, we get that $$g(n) = \frac{n(n-1)}{2} + 1$$ So $\boxed{g(2023) = \frac{2023\cdot 2022}{2} + 1 = 2045254}$