= Solution
The real <cube root> in the denominator vanishes at $x=-1$, so the printed rule needs the <domain of a function> $D_1=\mathbb R\setminus\{-1\}$. On its actual <domain of a function>, the second <iterated function> simplifies to
$$
f(f(x))=
\frac{x/\sqrt[3]{1+x^3}}{\sqrt[3]{1+x^3/(1+x^3)}}
=\frac{x}{\sqrt[3]{1+2x^3}}.
$$
This suggests
$$
\boxed{f^{(n)}(x)=\frac{x}{\sqrt[3]{1+nx^3}},\qquad
D_n=\mathbb R\setminus\{-j^{-1/3}:1\leq j\leq n\}.}
$$
For the induction step, assume the formula at $n$ and restrict to arguments for which its value is not $-1$. Its cube is $x^3/(1+nx^3)$, so another application gives
$$
f(f^{(n)}(x))
=\frac{x/\sqrt[3]{1+nx^3}}{\sqrt[3]{(1+(n+1)x^3)/(1+nx^3)}}
=\frac{x}{\sqrt[3]{1+(n+1)x^3}}.
$$
Real <cube roots> are multiplicative, including for negative arguments. The newly excluded point satisfies $x^3/(1+nx^3)=-1$, hence $x^3=-1/(n+1)$; all earlier exclusions remain. This proves both the formula and the <domain of a function> by <mathematical induction>. The inherited exclusions are essential for an <iteration of a partial function>: cancellation in the final expression cannot repair an undefined intermediate application.
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