((x^4y^3)^1/3(x^2y^2)^2-3)/x^1/3y^1/3

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Solution for ((x^4y^3)^1/3(x^2y^2)^2-3)/x^1/3y^1/3 equation: 


D( x )

x^1 = 0

x^1 = 0

x^1 = 0

x = 0

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

(y^1*((((((x^4*y^3)^1)/3)*(x^2*y^2)^2-3)/(x^1))/3))/3 = 0

(y*(((((x^4*y^3)/3)*(x^2*y^2)^2-3)/x)/3))/3 = 0

(y*(1/3*x^8*y^7-3))/(3^2*x) = 0

1/3*x^8*y^7 = 3 // : 1/3*y^7

x^8 = 9*y^-7

x^8 = 9*y^-7 // ^ 1/8

abs(x) = 9^(1/8)*y^(-7/8)

x = 9^(1/8)*y^(-7/8) or x = -(9^(1/8)*y^(-7/8))

x in { 9^(1/8)*y^(-7/8), -(9^(1/8)*y^(-7/8)) }

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