How to obtain a formula for $f(z)$ given this recurrence I am trying to figure out how to derive a formula for $f(z)$ that is a function of $z$ and maybe $k \in \mathbb{N}$:
$$
f(z) = 1+z f \bigg(\frac{z}{1+z} \bigg)
$$
As an attempt, I tried a change of variable $z=\frac{1}{x}$, and I get:
$$
f\bigg(\frac{1}{x}\bigg) = 1 + \frac{1}{x}f\bigg(\frac{1/x}{1+1/x}\bigg)
$$
which evaluates to the following after simplifying the terms inside $f$:
$$
f\bigg(\frac{1}{x}\bigg) = 1 + \frac{1}{x}f\bigg(\frac{1}{1+x}\bigg)
$$
Now, given the above, is the following telescoping operation valid?
$$
f\bigg(\frac{1}{x}\bigg) = 1 + \frac{1}{x}\bigg[ 1 + \frac{1}{x} + f\bigg(\frac{1}{2+x}\bigg) \bigg] = 1 + \frac{1}{x} + \frac{1}{(x+1)^2} + ... + \frac{1}{(x+k-3)^{k-2}} + \frac{1}{(x+k-2)^{k-1}}f\bigg(\frac{1}{k+x}\bigg)
$$
Also, how do I transform the above to a solution for $f(z)$?

If we plug in $z=0$ to the original functional equation we get $f(0)=1$. Then we set $g(x)=f(1/x)$. We have, as you showed, that $$g(x)=1+\frac1{x}g(x+1).$$
Thus for integer $m>0$ we have
$$g(x)=g(x+m+1)\prod_{r=0}^{m}\frac1{x+r}+\sum_{k=0}^{m-1}\prod_{j=0}^{k}\frac1{x+j}.$$
Taking the limit as $m\to\infty$ on both sides, we have $$g(x)=\sum_{k\ge0}\prod_{j=0}^{k}\frac1{x+j},$$
because $\lim_{m\to\infty}g(x+m)=\lim_{m\to\infty}f(\tfrac1{x+m})=f(\lim_{m\to\infty}\tfrac1{x+m})=f(0)=1$, while $\prod_{k\ge0}\frac1{x+k}=0$ for all $x\in\Bbb R\setminus \Bbb Z_{\le 0}$. Thus  we have
$$f(x)=\sum_{k\ge0}\prod_{j=0}^{k}\frac1{1/x+j}.$$
