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Partial answer : Mathematica's and Sage's algorithms give the same solutions :

sage: y=function("y")(x)
sage: SSol=desolve(6*diff(y,x)-2*y==x*y^4,y) ; SSol
e^(1/3*x)/(-1/2*(x - 1)*e^x + _C)^(1/3)
sage: MSol=mathematica("DSolve[6*y'[x]-2*y[x]==x*y[x]^4, y[x], x]") ; MSol
{{y[x] -> ((-2)^(1/3)*E^(x/3))/(-E^x + E^x*x - 2*C[1])^(1/3)}, 
 {y[x] -> -((2^(1/3)*E^(x/3))/(-E^x + E^x*x - 2*C[1])^(1/3))}, 
 {y[x] -> -(((-1)^(2/3)*2^(1/3)*E^(x/3))/(-E^x + E^x*x - 2*C[1])^(1/3))}}

Mathematica's answer is not (yet) easily translatable to Sage. However, one can use Mathematica to show that the three solutions given by Mathematica differ only by a multiplicative constant :

sage: KK=mathematica("E^(x/3)/(-E^x + E^x*x - 2*C[1])^(1/3)")
sage: [MSol[u][1][2]/KK for u in (1..3)]
[(-2)^(1/3), -2^(1/3), -((-1)^(2/3)*2^(1/3))]

Those three values are Mathematica's way of deboting the three cube roots of -2 :

sage: [mathematica.N(MSol[u][1][2]/KK) for u in (1..3)]
[0.6299605249474367 + 1.0911236359717214*I,
 0.6299605249474363 - 1.0911236359717216*I]
sage: [(MSol[u][1][2]/KK)^3 for u in (1..3)]
[-2, -2, -2]

With a bit of notational changes, the factor common to Mathematica's three solutions appears very close to the Sage's solution :

sage: C=function("C")
sage: var("_C")
sage: SSol.subs(_C==C(2))
e^(1/3*x)/(-1/2*(x - 1)*e^x + C(2))^(1/3)
sage: KK.sage(locals={"C":C,"E":e}).factor()
e^(1/3*x)/((x - 1)*e^x - 2*C(1))^(1/3)

up to a factor (-1/2)^(1/3) appearing in the denominator, which would be a factor (-2)^(1/3) in the numerator.

Sage appears to denote by (-1/2)^(1/3) any quantity whose cube is -1/2. Up to this notational quirk, Sane's and Mathematica's solutions are identical.