Answer: Option C
$$\eqalign{
& x = \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 - 1} }} \times \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 + 1} }} \cr
& \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{ {3 - 1} }} \cr
& \,\,\,\,\,\, = \frac{{3 + 1 + 2\sqrt 3 }}{2} \cr
& \,\,\,\,\,\, = 2 + \sqrt 3 \cr
& y = \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 + 1} }} \times \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 - 1} }} \cr
& \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{ {3 - 1} }} \cr
& \,\,\,\,\,\, = \frac{{3 + 1 - 2\sqrt 3 }}{2} \cr
& \,\,\,\,\,\, = 2 - \sqrt 3 \cr
& \therefore {x^2} + {y^2} = {\left( {2 + \sqrt 3 } \right)^2} + {\left( {2 - \sqrt 3 } \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {4 + 3} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr} $$