If you are trying to find the entries with numerical difference being zero, you might end up with nothing because the differences may be non-zero for all entries because they are double, and assuming these 616 entries are approximation to something.
I suggest you might wanna try to set a tolerance Tol: for example
X = sin(linspace(0,2*pi,616)');
% this gives you an X being the grid approximation to sine
Y = diff(X);
Tol = 1e-2*2*pi/616;615;
indY = find(abs(Y) < Tol);
% the index of the difference vector that has absolute value < Tolerance
Then X(indY) and X(indY+1) will give you the entries of X that produces these less-than-tolerance differences. If the latter is bigger than the former, then X starts to increase (approximation), vice versa.
MATLAB gives:
>> X(indY)
ans =
0.99997
1
-0.99992
-1
>> X(indY+1)
ans =
1
0.99992
-1
-0.99997
The criteria for choosing Tol is more like an a priori guess. First you have the length of the interval I set being $2\pi$, there is $616$ grid points, which has a magnitude of 1e2. The true derivative of $\sin x$ is $\cos x$, which has absolute value less than $1$, so the all the numerical differences should be less than or equal to $h = 2\pi/615$. Now I already know sine, when near its extrema, increases or decreases like a linear function (imagine Taylor expansion, or small angle approximation), and it takes about 100 grids to make it the maximum (knowing a priori for sine). Therefore the absolute value of the difference being less than the magnitude of the 1e-2 times the grid size would suffice to say that, X almost remains the same. It is problem dependent.