Just throw the "barriers" away!

Because to the condition $xy>0$ , both $x$ and $y$ must be both positive and negative and either case will lead to the same minimum positive value of the expression. Hence we can consider $x, y > 0$ and use AM-GM inequality as follows.

$\begin{aligned} \left(x + \frac 1x \right)\left(x + \frac 1y \right) + \left(y + \frac 1y \right)\left(y + \frac 1x \right) & = x^2 + \frac xy + 1 + \frac 1{xy} + y^2 + \frac yx + 1 + \frac 1{xy} \\ & = x^2 + y^2 + \frac 2{xy} + \frac xy + \frac yx + 2 \end{aligned}$

$\begin{aligned} {\color{#3D99F6}x^2 + y^2} + \frac 2{xy} + {\color{#D61F06}\frac xy + \frac yx} + 2 & \ge {\color{#3D99F6}2\sqrt{x^2y^2}} + \frac 2{xy} + {\color{#D61F06}2 \sqrt{\frac xy \cdot \frac yx}} + 2 \\ & \ge {\color{#3D99F6}2xy} + \frac 2{xy} + {\color{#D61F06}2} + 2 \\ & \ge 2\sqrt{2xy \cdot \frac 2{xy}} + 2 + 2 \\ & \ge 4 + 2 + 2 = \boxed{8} \end{aligned}$

Equality occurs when $x=y= \pm 1$ .

Note that $x>0,y>0$ or $x<0,y<0$ . If $x<0,y<0$ then let $a=-x,b=-y$ . Then $\large \left(x+ \frac 1x \right)\left(x+ \frac 1y\right)+\left(y+ \frac 1y \right)\left(y+ \frac 1x\right)$ $=$ $\large \left(a+ \frac 1a \right)\left(a+ \frac 1b\right)+\left(b+ \frac 1b \right)\left(b+ \frac 1a\right)$ , which is positive.

Observe that $\large \left(x+ \frac 1x \right)\left(x+ \frac 1y\right)+\left(y+ \frac 1y \right)\left(y+ \frac 1x\right) \ge 2(x+y+\frac { 1 }{ x } +\frac { 1 }{ y } )\ge 2(x+y+\frac { 4 }{ x+y } )\ge 8$ .

Equality holds when $x=y=\pm 1$ .