Answer: Option D
$$\eqalign{
& {\text{Given,}} \cr
& x = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}{\text{ , }}y = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& {\text{Find, }}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = ? \cr
& \Rightarrow {\text{ }}\frac{{{x^2} + {y^2} + 2xy - xy}}{{{x^2} + {y^2} - 2xy + xy}} \cr
& \Rightarrow {\text{ }}\frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} = ? \cr
& {\text{Now,}} \cr
& {\text{ }}x + y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& \Rightarrow x + y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2} + {{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{{{\sqrt 5 }^2} - {{\sqrt 3 }^2}}} \cr
& \Rightarrow {\text{ }}x + y = \frac{{2\left( {{{\sqrt 5 }^2} + {{\sqrt 3 }^2}} \right)}}{{5 - 3}} \cr
& \Rightarrow x + y = 8\,.......(i) \cr
& Again, \cr
& x - y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& \Rightarrow {\text{ }}x - y = \frac{{4 \times \sqrt 5 \times \sqrt 3 }}{2} \cr
& \Rightarrow x - y = 2\sqrt {15} ..............(ii) \cr
& {\text{And, }}xy = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& \Rightarrow {\text{ }}xy = 1 \cr
& {\text{Substitutes values in the question}}{\text{.}} \cr
& \Rightarrow \frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} \cr
& \Rightarrow \frac{{{8^2} - 1}}{{{{\left( {2\sqrt {15} } \right)}^2} + 1}} \cr
& \Rightarrow \frac{{63}}{{61}} \cr} $$