The greatest value of  $f\left(x\right)=(x+1{)}^{1/3}-(x-1{)}^{1/3}$  on [0, 1] is

$\Rightarrow f^1\left(x\right)=\frac{(x-1{)}^{\frac{2}{3}}-(x+1{)}^{\frac{2}{3}}}{3{(x^2-1)}^{\frac{2}{3}}}$

clearly $f^1\left(x\right)$ does not exists at $x=\pm 1$.

Now $f^1\left(x\right)=0$  then $x=0$

clearly $f^1\left(x\right)\ne 0$  for any other value of $x=\pm 1$

The value of $f\left(x\right)$  at $x=0$ is 2

Hence the greatest value of $f\left(x\right)$  is 2