Wolfram Alpha produces this plot for x <= y, xy >= 1, y <= 2
:

So it makes sense that you would need to break the integration up into at least two pieces. Since $xy > 0$, $x$ and $y$ are always of the same sign.
When $x > 0, y> 0$, then $y = 2$ is the top, $y = x$ is bottom and left, and $y = 1/x$ is bottom and right. $y = 1/x$ and $y = x$ cross at $(1,1)$, so if you integrate by $y$ first, you need $y \in [1/x, 2]$ for $x \le 1$ and $y \in [x, 2]$ for $x \ge 1$. $y = 1/x$ crosses $y = 2$ at $x = 1/2$, so you get $\int_{1/2}^1\int_{1/x}^2 \, dydx$ for the first and $\int_1^2\int_{x}^2 \, dydx$ for the second.
But if you integrate $x$ first, then this piece can be done in a single integration: $\int_1^2\int_{1/y}^y \, dxdy$
The lower left piece where $x < 0, y < 0$ is simpler, as $y \le 1/x$ implies $y \le 2$ here. The integral is $\int_{-\infty}^{-1}\int_{x}^{1/x}\, dydx$
So the simplest form is
$$\int_E \frac{y^2}{x^2}d\lambda^2(x,y) = \int_{-\infty}^{-1}\int_{x}^{1/x}\frac{y^2}{x^2}\,dydx + \int_1^2\int_{1/y}^y \frac{y^2}{x^2}\, dxdy$$