We see that we need to find $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}$, which simplifies to $\frac{a+b+c}{abc}.$ From here, we can now proceed with Vieta's. By Vieta's theorem (if we assume that the polynomial $f(x) = Ax^3 + Bx^2 + Cx + D$, we know that a+b+c=−BA, ab+bc+ac=CA, abc=−DA, where $A=1, B=-6, C=13, D=-18.$ Therefore, we get that $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} = \frac{-6}{-18} = \boxed{\frac{1}{3}}.$