What is the function $f(x)$ which is differentiable everywhere and $f(x-1)f(x-2)+1=f(x)$?

Inspired by Milten's comment, here's some dirty asymptotic calculation:

1. Approximate the original equation $$f(x) \approx f(x-1)f(x-2)$$ Given the terms you have written, this approximation should be pretty good everywhere except for the first few terms

2. Let $g(x) = \log f(x)$. Then

$$g(x) = g(x-1) + g(x-2)$$

is a Fibonacci sequence. The explicit form of this sequence is given by

$$g(x) = \frac{\phi^x-\psi^x}{\sqrt5}$$

where $\phi$ and $\psi$ are golden ratio and its square-conjugate. Since $\psi\approx-0.6$, it will asymptotically tend to zero, so the asymptotic behaviour of the approximated sequence will be

$$f(x) \approx \exp \frac{\phi^x}{\sqrt 5}$$

Probably this approach can be cleaned up by considering very precisely the first few initial values of the sequence $g(x)$ to obtain expressions precise to a desired order. Analytic closed-form expression likely does not exist, because the recursive identity combines both addition and multiplication.

I will prove the

Statement

There is no continuous function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that it satisfies following conditions:

1. $\exists x \in \mathbb{R}$ s.t. $f\left(x\right) = 0$;
2. $\forall x \in \mathbb{R}$ $$ f\left(x - 1\right)f\left(x\right) + 1 = f\left(x + 1\right)$$

Proof

Assume that there is such $f$. Let $\mathbb{U}$ be the set of zeros of $f$. $\mathbb{U}$ is nonemtpy. Let $u \in \mathbb{U}$. Then we have that $$ f\left(u - 2\right)f\left(u - 1\right) = -1$$ Meaning that both $f\left(u - 2\right)$ and $f\left(u - 1\right)$ are nonzero and have different signs. By continuity there exists $$\mathbb{U} \ni u' \in \left(u - 2, u - 1\right)$$ Thus we have shown that for any $x \in \mathbb{R}$ there is $u \in \mathbb{U}$ such $u < x$.

If $u \in \mathbb{U}$, then we have the following equalities: $$\begin{align*} f\left(u + 0\right) &= 0 \\ f\left(u + 1\right) &= 1 \\ f\left(u + 2\right) &= 1 \\ f\left(u + 3\right) &= 2 \\ f\left(u + n\right) &= a_{n + 1},\quad n \in \mathbb{N} \tag{1} \label{u+n} \end{align*}$$ , where $a_{n}$ defined here. From this sequence we only need to know that $a_{n}$ is non-decreasing and $$ \lim\limits_{n \rightarrow \infty} a_{n} = +\infty \tag{2} \label{a_n}$$ By the Weierstrass theorem $f$ reaches its maximum $\mathsf{M} \in \mathbb{R}$ on $\left[0, 1\right]$. By \eqref{a_n} we can find $n \in \mathbb{N}$ such that $a_{n} > \mathsf{M}$. Then we find $u \in \mathbb{U}$ and $m \in \mathbb{N}$ s.t. $-m \leq u \leq -m + 1 \leq -n$. It follows that $u + m \in \left[0, 1\right]$ and $m > n$. But by \eqref{u+n} $$\begin{align*} \mathsf{M} &\geq f\left(u + m\right) \\ &\geq a_{m + 1} \\ &\geq a_{n} \\ &> \mathsf{M} \end{align*}$$ This completes the proof.

It follows that there is no function you are looking for, if it's defined like $f : \mathbb{R} \rightarrow \mathbb{R}$.

P. S. English is not my native language. I apologize for any possible mistakes when writing the answer in English.