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

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


x in (-oo:+oo)

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

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

3*(y^3-3*x) = 0

( 3 )

3 = 0

x belongs to the empty set

( y^3-3*x )

y^3-3*x = 0 // - y^3

-3*x = -y^3 // : -3

x = (-y^3)/(-3)

x = (y^3)/3

x = (y^3)/3

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