Step 1 :The maximum number of real zeros a polynomial can have is equal to its degree. In this case, the degree of the polynomial is 4, so the maximum number of real zeros is 4.
Step 2 :Descartes' Rule of Signs states that the number of positive real zeros of a polynomial is equal to the number of sign changes in the polynomial's terms, or less than that by an even number. The polynomial \(f(x)=6 x^{4}-8 x^{2}-x+9\) has 2 sign changes (from positive to negative at \(-8x^2\) and from negative to positive at \(9\)), so it can have 2 or 0 positive real zeros.
Step 3 :To find the number of negative real zeros, we substitute \(-x\) for \(x\) in the polynomial and then apply Descartes' Rule of Signs again. The polynomial becomes \(f(-x)=6 x^{4}-8 x^{2}+x+9\), which has 1 sign change (from negative to positive at \(x\)), so it can have 1 or 0 negative real zeros.
Step 4 :Final Answer: The maximum number of zeros that \(f(x)=6 x^{4}-8 x^{2}-x+9\) can have is \(\boxed{4}\). The function \(f(x)=6 x^{4}-8 x^{2}-x+9\) may have \(\boxed{2}\) or \(\boxed{0}\) positive real zeros. The function \(f(x)=6 x^{4}-8 x^{2}-x+9\) may have \(\boxed{1}\) or \(\boxed{0}\) negative real zeros.