Common Region of Equations:
To find the area enclosed by the common region of the given equations, we need to determine the points of intersection between the two curves.

Determining the Points of Intersection:
We start by equating the two equations:

x^2 + y^2 = 2 ...(1)
2x - 3y = 0 ...(2)

By solving equations (1) and (2) simultaneously, we can find the points of intersection.

Solving the Equations:
From equation (2), we can express x in terms of y:

2x = 3y
x = (3y)/2

Substituting the value of x in equation (1), we get:

((3y)/2)^2 + y^2 = 2
(9y^2)/4 + y^2 = 2
(9y^2 + 4y^2)/4 = 2
13y^2 = 8
y^2 = 8/13
y = ±√(8/13)

Therefore, the points of intersection are (3√(8/13)/2, √(8/13)) and (-3√(8/13)/2, -√(8/13)).

Calculating the Area:
To calculate the area, we need to find the difference between the y-coordinates of the points of intersection and integrate the equation over that range.

Let's integrate equation (2) with respect to x:

∫(2x - 3y) dx = ∫0 dx
x^2 - 3yx = C

Substituting the points of intersection, we get:

(3√(8/13)/2)^2 - 3(√(8/13))(√(8/13)) = C
(9(8/13)/4) - 3(8/13) = C
(9(8/13) - 12(8/13))/4 = C
(72/13 - 96/13)/4 = C
(-24/13)/4 = C
C = -6/13

Thus, the equation becomes:

x^2 - 3yx = -6/13

Calculating the Area Enclosed:
To find the area enclosed, we integrate the equation over the range of y-coordinates from -√(8/13) to √(8/13):

A = ∫[-√(8/13), √(8/13)] (x^2 - 3yx) dy

Simplifying the integral:

A = ∫[-√(8/13), √(8/13)] (-6/13) dy
A = (-6/13) * [y] [-√(8/13), √(8/13)]
A = (-6/13) * (√(8/13) - (-√(8/13)))
A = (-6/13) * 2√(8/13)
A = -12√(8/13)/13

Therefore, the area enclosed by the common region of the given equations is -12√(8/13)/13 square units.