The identity involves going through real numbers to get the integer result, and if you're using floating point or fixed point calculations, you need a good amount of accuracy to get the correct integer out, which also won't be the fastest thing. Using https://en.wikipedia.org/wiki/Fibonacci_number#Matrix_form you can get a different way to calculate Fibonacci numbers recursively, which involves far fewer computations.
Fundamentally, the thing to do is to understand the N-th Fibonacci number as a + b, where φ^N = a + b * φ, where a and b are natural numbers and φ^2 = φ + 1.
It's not important to understand phi as a floating point value; we can just work in the abstract arithmetic where we've added to the natural numbers a new entity phi defined to satisfy φ^2 = φ + 1 (just like complex numbers are the abstract arithmetic where we've added to ordinary arithmetic an i defined to satisfy i^2 = -1).
Call these Fomplex numbers; a Fomplex number a + b * φ amounts to just a pair of natural numbers, and it's easy enough to add and multiply them with natural number arithmetic. To calculate the N-th Fibonacci number, just calculate φ^N and add together its coefficients, as noted. As for how to efficiently calculate φ^N, use the usual addition chain approach to exponentiation (e.g., "repeated squaring"), thus getting a result in Θ(log N) many additions and multiplications.
This is incidentally the same as the matrix approach, essentially, but perhaps a cleaner perspective on it; at any rate, it is a way of thinking which will serve as a useful tool in your back pocket for other general problems about linear recurrences.
How neat! Among its properties not the least interesting is that it has finally convinced me that the initial condition f(0) = f(1) = 1 is not arbitrary but perfectly natural.
Ah, but actually that part is still a little arbitrary. This technique is fully general; variants describe ANY linear recurrence with ANY initial values. In particular, the Nth value of the Fibonacci-type sequence with 0th value x and 1th value y is xa + yb where a + bϕ = ϕ^N. I just happen to have chosen the weights x and y to both be 1 in that discussion.
