Let $g\left(x\right)$ be the inverse of the function $f\left(x\right)$ and $f^1\left(x\right)=\frac{1}{1+x^3}.$ Then $g^1\left(x\right)$ is equal to

Since $g\left(x\right)$ is the inverse of $f\left(x\right)$, therefore $f\left(x\right)=y\iff g\left(y\right)=x$

Now, $g^{\mathrm{\prime }}\left(f\right(x\left)\right)=\frac{1}{f^{\mathrm{\prime }}\left(x\right)}, \mathrm{\forall }x\Rightarrow g^{\mathrm{\prime }}\left(f\right(x\left)\right)=1+x^3, \mathrm{\forall }x$ $\Rightarrow g^{\mathrm{\prime }}\left(y\right)=1+\left(g\right(y){)}^3$

[using $f\left(x\right)=y\iff x=g\left(y\right)$] $\Rightarrow g^{\mathrm{\prime }}\left(x\right)=1+\left(g\right(x){)}^3$ (replacing y by x)