A functional equation

Prove that this equation does not admit any real continuous function R ->R ( try Rⁿ->R^m)

f(x).f(x+1) + f(x) + 1 = 0

My progress so far: turn this into a recursive equation
f(x+1)= -1/ [ f(x) +1]
suppose if f(x_o)=k
then , f(x_o+4)=k for arbitrary x_o and k belonging to R.
Let us select x and x+$ , two real numbers, then
epsilon=f(x)-f(x+$) = $/ (1+x)²
//neglecting $ as it does not matter in the analysis that follows//
if x is negative and greater than -1, epsilon> $.
will this help? am not able to proceed from here .