Consider the first equation i.e

$9xy = 4 ,i.e., 9x = 4/y$

Let $x = 0$ then $y = ∞$

So, $(∞, 0)$

When $x = 1 ,y = 4/9$

When $y = 1 ,x = 4/9$

Plotting rough diagram

We get

Now consider 2nd equation

i.e $2x+y = 2 \to(2)$

let $x = 0 ,y = 2$

let $y = 0 ,x= 1$

Equation (2) is a.

Straight line passing from $(0, 2)$ and $(1, 0)$

This line is intersecting curve at

Solving eq (2)

$xy = 4/9 ,x = 4/9y$

Substituting value of x in eq (2)

$\therefore \dfrac {2\times 4}{9y}+y=2\\ 8+9y^2=18y $

$ \therefore y=\dfrac 43 \space and \space \dfrac 23$

$\text { Corresponding x is }$

$ x=\dfrac 4{9\times \frac 43}=\dfrac 13$

$x=\dfrac 4{9\times \frac 23}=\dfrac 23 $

Hence the points of intersections of the hyperbola and the line are $(2/3, 2/3)$ and $(1/3, 4/3)$

Now considering the horizontal strip which will slide along x axis i.e from $1/3$ to $2/3$ which is Outer limit

Upper limit is equation of curves $9xy = 4$ in terms of y

We get $y =\dfrac 4{9x}\\ \therefore A=\int\limits_{1/3}^{2/3}\int\limits_{4/9x}^{2-2x}dydx$

Integrating w.r.t y we get

$A=\int\limits_{1/3}^{2/3}[y]^{2-2x}_{4/9x}dx \\ A= \int\limits_{1/3}^{2/3}2-2x- \dfrac 4{9x}dx \\ A=\Bigg[2x-\dfrac {2x^2}2-\dfrac 49 \log x\Bigg]_{1/3}^{2/3}\\ =\dfrac 43-\dfrac 49-\dfrac 49\log \dfrac 23\\ =\dfrac 23+\dfrac 19+\dfrac 49\log\dfrac 13 \\ =\dfrac 13-\dfrac 49[\log 2-\log 3-\log 1+\log 3]\\ =\dfrac 13-\dfrac 49\log 2$