Solution:
Since square numbers are always non-negative we have
$$(x - y)^{2} \geq 0 \qquad \mathrm{and} \qquad (x y - 2)^{2} \geq 0.$$
Also since $x$ and $y$ are positive we have $\frac{2}{x y} > 0$. Combining this all together gives us:
$$(x - y)^{2} + \frac{2}{x y} (x y - 2)^{2} \geq 0.$$
From here we expand and simplify:
$$(x^{2} - 2x y + y^{2}) + \frac{2}{x y} (x^{2}y^{2} - 4x y + 4) \geq 0$$
$$x^{2} - 2x y + y^{2} + 2x y - 8 + \frac{8}{x y} \geq 0$$
$$x^{2} + \frac{8}{x y} + y^{2} \geq 8$$
as required.

Alternative Solution:
Consider the AM-GM inequality applied to $\left\{x^{2}, \frac{4}{x y}, \frac{4}{x y}, y^{2}\right\}$
$$\frac{x^{2} + \frac{4}{x y} + \frac{4}{x y} + y^{2}}{4} \geq \sqrt[4]{x^{2} \times \frac{4}{x y} \times \frac{4}{x y} \times y^{2}}$$
$$\frac{x^{2} + \frac{8}{x y} + y^{2}}{4} \geq 2$$
$$x^{2} + \frac{8}{x y} + y^{2} \geq 8.$$