Inspired from dev sharma's note

Take that one equation is the inverse of the other, thus they must find each other in a point (x, x). Then we can write x^6 = x^4 + 18 => x^6 - x^4 - 18 = 0. Let a = x^2, we have a^3 - a^2 - 18 = 0, which is easy to solve and get a = 3, as a must be real. Finally, as x^2 = 3 and x = y, the answer is 12.

$\begin{cases} x^6=y^4+18 \implies y^4 = x^6-18 & \implies y^{12} = x^{18} -54x^{12} + 972 x^6 -5832 \\ y^6=x^4+18 & \implies y^{12} = x^8 +36x^4 + 324 \end{cases}$

Then we have:

$\begin{aligned} x^{18} -54x^{12} + 972 x^6 - 5832 & = x^8 +36x^4 + 324 \\ x^{18} -54x^{12} - x^8 + 972 x^6 - 36x^4 - 6156 & = 0 & \small \color{#3D99F6} \text{Let }u = x^2 \\ u^9 -54u^6 - u^4 + 972 u^3 - 36u^2 - 6156 & = 0 & \small \color{#3D99F6} \text{By rational root theorem} \\ (u-3)(u^8+3u^7+9u^6-27u^5-81u^4-244u^3+240u^2+684u+2052) & = 0 \end{aligned}$

$u=3$ is the only real root. Therefore, $\begin{cases} x = \sqrt 3 & \implies \begin{cases} y = \sqrt 3 & (x,y) = (\sqrt 3, \sqrt 3) \\ y = - \sqrt 3 & (x,y) = (\sqrt 3, - \sqrt 3) \end{cases} \\ x = - \sqrt 3 & \implies \begin{cases} y = \sqrt 3 & (x,y) = (-\sqrt 3, \sqrt 3) \\ y = - \sqrt 3 & (x,y) = (-\sqrt 3, - \sqrt 3) \end{cases} \end{cases}$

Therefore, the answer is $(\sqrt 3)^2 + (-\sqrt 3)^2 + \sqrt 3)^2 + (-\sqrt 3)^2 = \boxed {12}$ .