Answer: Option C
$$\eqalign{
& {\text{Given, }}x + \frac{1}{x} = \sqrt {13} {\text{ }} \cr
& {\text{then ,}}\frac{{3x}}{{\left( {{x^2} - 1} \right)}} \cr
& = \frac{3}{{x - \frac{1}{x}}}\,.............(i) \cr
& {\text{Now, }}x + \frac{1}{x} = \sqrt {13} \cr
& {\text{On squaring both side}} \cr
& = {x^2} + \frac{1}{{{x^2}}} \cr
& = 13 - 2 \cr
& = 11 \cr
& = {x^2} + \frac{1}{{{x^2}}} - 2 \cr
& = 11 - 2 \cr
& = 9 \cr
& \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 9 \cr
& \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = {3^2} \cr
& \Rightarrow x - \frac{1}{x} = 3 \cr
& {\text{Put this value in equation (i)}} \cr
& \Rightarrow \frac{3}{{x - \frac{1}{x}}} \cr
& \Rightarrow \frac{3}{3} \cr
& \Rightarrow 1 \cr} $$