Is this a new Fibonacci Identity? I have found the following Fibonacci Identity (and proved it).
If $F_n$ denotes the nth Fibonacci Number, we have the following identity
\begin{equation}
    F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}
\end{equation}
where $F_1 = F_2 = 1$, $r \leq n$, $h \leq g$, and $n, g, k \in \mathbb{N}$. 
It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.
I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before?  I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!
 A: "Vajda's identity" is really Tagiuri's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12.
See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.
A: This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4\times 4$ matrix. 
Concrete mathematics gives the following reference: 

As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_{m,n}=s_{m+n}s_{m-n}$ $(m,n\in \mathbb{Z})$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.
A: Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $n\in {\mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.
