Here limit of y are $y=x$ to $y=1/x$ ->Hyperbola

And the limit of x is $x=0$ to $x=1$

Now the point of intersection of line $y=x$ & hyperbola is

Putting $y=x$ in $y=1/x$

$$\therefore x=1/x$$

$$\therefore x^2 = 1$$

$$\therefore x=1 $$

When $x=1 ,y=1$

$\therefore$ point if intersection is $(1,1)$

Now changing the order the region is divided into two parts as upper limits changes at $y=1$

$\therefore$ The horizontal strip will slide from $y=0$ to $y=∞$

Region (1)

1) Outer limit y

$Y=1$ to $y=∞ $

2) Inner limit x

a) Upper limit equation of hyperbola $x=1/y$

b) lower limit

$$\therefore I= \int\limits_0^1\int\limits_0^y\dfrac y{(1+xy)^2(1+y^2)}dxdy + \int\limits_1^{\infty}\int\limits_0^{\frac 14}\dfrac y{(1+xy)^2(1+y^2)}dxdy$$

$\therefore I=\int\limits_0^1\dfrac y{1+y^2} \int\limits_0^y(1+xy)^{-2}dxdy +\int\limits_1^{\infty}\dfrac y{1+y^2}\int\limits_0^{\frac 1y} (1+xy)^{-2}dxdy\\ \therefore I=\int\limits_0^1\dfrac y{1+y^2}\Bigg[\dfrac {[1+xy]^{-1}}{-1\times y}\Bigg]^y_0dy+\int\limits_1^{\infty}\dfrac y{1+y^2}\Bigg[\dfrac {[1+xy]^{-1}}{-1\times y}\Bigg]^{\frac 1y}_0dy\\ \therefore I= \int\limits_0^1 \dfrac y{1+y^2}\Bigg[\dfrac {(1+y^2)^{-1}}{-y}-\Bigg[\dfrac {-1}y\Bigg]\Bigg]dy + \int\limits^{\infty}_1 \dfrac y{1+y^2}\Bigg[\dfrac {(1+1)^{-1}}{-y}-\Bigg[\dfrac {-1}y\Bigg]\Bigg]dy\\ I=\int\limits^{11}_0 \dfrac y{1+y^2}\times \dfrac 1y\Bigg[1-\dfrac 1{1+y^2}\Bigg]dy + \int\limits_1^{\infty}\dfrac y{1+y^2}\times 1y\Bigg[1-\dfrac 12\Bigg]dy\\ I= \int\limits^{1}_0 \dfrac y{1+y^2}\times \dfrac {1+y^2-1}{1+y^2}dy+\int\limits^{\infty}_1 \dfrac y{1+y^2}\times \dfrac 12 dy\\ I=\int\limits_0^1\dfrac {y^2}{(1+y^2)^2}dy+\dfrac 12\int\limits_1^{infty}\dfrac 1{1+y^2}dy\\ Let \space I=I_1+I_2\\ I_1=\int\limits^1_0 \dfrac {y^2}{(1+y^2)^2}dy\\ Let \space y=\tan\theta \\ dy=\sec^2\theta\space d\theta\\ when \space y=0,\theta =0\\ y=1,\theta=\dfrac \pi4\\ \therefore I_1=\int\limits_0^{\frac \pi4}\dfrac {\tan^2\theta}{(1+\tan^2\theta)}\sec^2theta \space d\theta\\ I_1= \int\limits_0^{\frac \pi4}\dfrac {\tan^2\theta}{(\sec^2\theta)}\times\sec^2\theta d\sec^2\theta\\ I_1=\int\limits_0^{\frac \pi4}(1-\cos^2\theta)d\theta\\ =[\theta]^{\frac \pi4}_0- \int\limits_0^{\frac \pi4} \dfrac {1+\cos\theta}2d\theta\\ =\dfrac \pi4-\dfrac 12[\theta+\dfrac {\sin2\theta}2]_0^{\frac \pi4}\\ =\dfrac \pi4-\dfrac 12[\dfrac \pi4+\dfrac -0-0]\\ =\dfrac \pi8-\dfrac 14\\ I_2=\dfrac 12\int\limits_1^{\infty}\dfrac 1{1+y^2}dy\\ =\dfrac 12[\tan^{-1}y]_1^{\infty}\\ =\dfrac 12[\tan^{-1}\infty-\tan^{-1}1] \\ =\dfrac 12{\dfrac \pi2-\dfrac \pi4}]\\ =\dfrac \pi8\\ \therefore I=I_1+I_2\\ \therefore I=\dfrac \pi8-\dfrac 14+\dfrac \pi8\\ =\dfrac \pi4-\dfrac 14\\ =\dfrac {\pi-1}4$