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?

You got $$y'=-\dfrac{2x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)^\frac23$$
You can leave till here if you want. If you want to match the given answer, multiply and divide by $\left(\dfrac{1+x^3}{1-x^3}\right)^\frac13$, thus,
$$y'=-\dfrac{2x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)\left(\dfrac{1-x^3}{1+x^3}\right)^\frac13\\=\dfrac{2x^2}{x^6-1}\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$
If you want to reach the given answer directly, you can take $\log$ on the given expression, thus,
$$\log y=\frac13\left(\log(1-x^3)+\log(1+x^3)\right)$$
Taking derivative,
$$\frac{y'}{y}=\frac13\left(\frac{-3x^2}{1-x^3}-\frac{3x^2}{1+x^3}\right)\\=x^2\left(\frac{-1-x^3-1+x^3}{1-x^6}\right)$$
Therefore, $$y'=\dfrac{2x^2}{x^6-1}\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$
