Find the first derivative of $y=\sqrt[3]{\frac{1-x^3}{1+x^3}}$ Find the first derivative of $$y=\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$
The given answer is $$\dfrac{2x^2}{x^6-1}\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$ It is nice and neat, but I am really struggling to write the result exactly in this form. We have $$y'=\dfrac13\left(\dfrac{1-x^3}{1+x^3}\right)^{-\frac23}\left(\dfrac{1-x^3}{1+x^3}\right)'$$ The derivative of the "inner" function (the last term in $y'$) is $$\dfrac{-3x^2(1+x^3)-3x^2(1-x^3)}{\left(1+x^3\right)^2}=\dfrac{-6x^2}{(1+x^3)^2},$$ so for $y'$ $$y'=-\dfrac13\dfrac{6x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)^\frac23=-\dfrac{2x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)^\frac23$$ Can we actually leave the answer this way?

HINT
I would start with noticing that
\begin{align*}
y^{3} = \frac{1 - x^{3}}{1 + x^{3}} = -1 + \frac{2}{1 + x^{3}} \Rightarrow 3y^{2}y' = -\frac{6x^{2}}{(1 + x^{3})^{2}}
\end{align*}
Can you take it from here?
