Given equation is 5x^2 + 6xy + 2y^2 + 7x + 6y + 6 = 0. To find the area enclosed by the closed curve, we need to convert the given equation into standard form of equation of an ellipse.

Steps to convert the equation into standard form:
1. Collect the terms with x and y together.
5x^2 + 6xy + 2y^2 + 7x + 6y + 6 = 0
5x^2 + (6y + 7)x + (2y^2 + 6y + 6) = 0

2. Solve for x using quadratic formula.
x = [-6y - 7 ± √(36y^2 + 84y + 49 - 40y^2 - 48y - 24)] / 10
x = [-6y - 7 ± √(-4y^2 - 12y + 25)] / 10
x = [-6y - 7 ± √(4y + 5)(-y - 5)] / 10

3. Substitute the value of x in the original equation.
5x^2 + 6xy + 2y^2 + 7x + 6y + 6 = 0
5[-6y - 7 ± √(4y + 5)(-y - 5)]^2 + 6y[-6y - 7 ± √(4y + 5)(-y - 5)] + 2y^2 + 7[-6y - 7 ± √(4y + 5)(-y - 5)] + 6y + 6 = 0

4. Simplify the equation and group the terms.
[(4y + 5)(-y - 5)]y^2 + [60(4y + 5) - 36(-y - 5) - 49]y - 35(4y + 5) = 0
-4y^4 - 46y^3 - 78y^2 - 60y + 175 = 0

5. Divide the equation by -4 to get standard form of equation of an ellipse.
y^2 + (23/2)y + (39/8)x^2 + (15/4)x - (175/4) = 0
[(x + 5/13)^2 / (39/52)] + [(y + 23/52)^2 / (39/208)] = 1

The equation is in standard form of equation of an ellipse with center (-5/13, -23/52), semi-major axis √(39/52) and semi-minor axis √(39/208).

The area enclosed by the closed curve is π times the product of the semi-major axis and semi-minor axis.
Area = π(√(39/52))(√(39/208))
Area = π(3/4)
Area = 3π/4

Therefore, the correct option is C) /2.