Differentiation of $\frac{x^3}{1-x^3}$ with respect to $x^3$ is equal to:

$\frac{1}{(1-x^3)^2},x≠1$

The correct answer is Option (1) → $\frac{1}{(1-x^3)^2},x≠1$ **

Let $u = x^3$.

Given expression: $\frac{x^3}{1 - x^3} = \frac{u}{1 - u}$

Differentiate w.r.t. $u$:

$\frac{d}{du}\left(\frac{u}{1-u}\right) = \frac{(1-u)\cdot 1 - u(-1)}{(1-u)^2}$

$= \frac{1 - u + u}{(1-u)^2}$

$= \frac{1}{(1-u)^2}$

Substitute $u = x^3$:

$\frac{1}{(1 - x^3)^2}$

The differentiation with respect to $x^3$ equals $\frac{1}{(1-x^3)^2}$.