(3x^3/2y^3/x^2y^-1/2)^-2

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Solution for (3x^3/2y^3/x^2y^-1/2)^-2 equation: 


D( x )

(((((3*x^3)/2)*y^3)/(x^2))*y^-1)/2 = 0

x^2 = 0

(((((3*x^3)/2)*y^3)/(x^2))*y^-1)/2 = 0

(((((3*x^3)/2)*y^3)/(x^2))*y^-1)/2 = 0

3/4*x*y^2 = 0 // : 3/4*y^2

x = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

x in (-oo:0) U (0:+oo)

((((((3*x^3)/2)*y^3)/(x^2))*y^-1)/2)^-2 = 0

16/9*x^-2*y^-4 = 0 // : 16/9*y^-4

x^-2 = 0

x naleu017Cy do O

x belongs to the empty set

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