Answer: Option C
$$\eqalign{
& x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ and y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr
& \therefore x = \frac{1}{y} \cr
& \Leftrightarrow xy = 1 \cr
& x + y = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr
& \Rightarrow x + y = \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}} \cr
& \Rightarrow x + y = \frac{{12}}{4} \cr
& \Rightarrow x + y = 3 \cr
& \Rightarrow x + \frac{1}{x} = 3 \cr
& \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {\left( 3 \right)^2} - 2 \cr
& \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr
& {\text{Now,}}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} \cr
& = \frac{{{x^2} + {y^2} + xy}}{{{x^2} + {y^2} - xy}} \cr
& = \frac{{7 + 1}}{{7 - 1}} \cr
& = \frac{8}{6} \cr
& = \frac{4}{3} \cr} $$