这道题怎么求导，求详细过程
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y = [(1+x^3)/(1-x^3)]^(1/3),
lny = (1/3)[ln(1+x^3)-ln(1-x^3)]
y'/y = (1/3)[3x^2/(1+x^3) - (-3x^2)/(1-x^3)] = x^2/(1+x^3) + x^2/(1-x^3)
y' = [(1+x^3)/(1-x^3)]^(1/3) [x^2/(1+x^3) + x^2/(1-x^3)]
lny = (1/3)[ln(1+x^3)-ln(1-x^3)]
y'/y = (1/3)[3x^2/(1+x^3) - (-3x^2)/(1-x^3)] = x^2/(1+x^3) + x^2/(1-x^3)
y' = [(1+x^3)/(1-x^3)]^(1/3) [x^2/(1+x^3) + x^2/(1-x^3)]
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