A regular pyramid

Let the length of the slanting side of the isosceles triangle be $a$ . Then $a=\sqrt{15^2 + 6^2} = \sqrt{261} = 3\sqrt{29}$ . By Heron's formula , the area of an isosceles triangle is:

$\begin{aligned} A_\triangle & = \sqrt{(a+3)(3^2)(a-3)} & \small \blue{\text{Note that }s = \frac {2a+6}2 = a+3} \\ & = \sqrt{9(a^2 - 9)} = \sqrt{9\cdot 9 \cdot 28} \\ & = 18 \sqrt 7 \end{aligned}$

The area of the base hexagon is that of six equilateral triangles with side length $6$ . Then $A_{\text{hex}} = 6 \cdot \dfrac {6^2 \sin 60^\circ}2 = 54 \sqrt 3$ .

Therefore the total surface area $A = A_{\text{hex}} + 6 A_\triangle = 54\sqrt 3 + 108 \sqrt 7$ . $\implies x+y = 54+108 = \boxed{162}$ .

Area of the regular hexagonal base: $A_{1} = 6 \cdot (\frac{6^2 \sqrt{3}}{4}) = 54\sqrt{3}.$

Area of the six slanted faces: $A_{2} = 6 \cdot (\frac{1}{2} \cdot 6 \cdot \sqrt{15^2 + (3\sqrt{3})^2}) =108\sqrt{7}.$

Hence, $x=54, y=108 \Rightarrow x+y = \boxed{162}.$