(a) Since [tex]g(x)=\sqrt[3]{x}[/tex] and [tex]h(x) = \frac1{x^3}[/tex], we have
[tex](g\circ h)(x) = g(h(x)) = g\left(\dfrac1{x^3}\right) = \sqrt{3}{\dfrac1{x^3}} = \dfrac1x[/tex]
We're given that
[tex](f \circ g \circ h)(x) = f(g(h(x))) = f\left(\dfrac1x\right) = \dfrac x{x+1}[/tex]
but we can rewrite this as
[tex]\dfrac x{x+1} = \dfrac{\frac xx}{\frac xx + \frac1x} = \dfrac1{1+\frac1x}[/tex]
(bear in mind that we can only do this so long as x ≠ 0) so it follows that
[tex]f\left(\dfrac1x\right) = \dfrac1{1+\frac1x} \implies \boxed{f(x) = \dfrac1{1+x}}[/tex]
(b) On its own, we may be tempted to conclude that the domain of [tex](f\circ g\circ h)(x) = \frac1{1+x}[/tex] is simply x ≠ -1. But we should be more careful. The domain of a composite depends on each of the component functions involved.
[tex]g(x) = \sqrt[3]{x}[/tex] is defined for all x - no issue here.
[tex]h(x) = \frac1{x^3}[/tex] is defined for all x ≠ 0. Then [tex](g\circ h)(x) = \frac1x[/tex] also has a domain of x ≠ 0.
[tex]f(x) = \frac1{1+x}[/tex] is defined for all x ≠ -1, but
[tex](f\circ g\circ h)(x)=f\left(\frac1x\right) = \dfrac1{1+\frac1x}[/tex]
is undefined not only at x = -1, but also at x = 0. So the domain of [tex](f\circ g\circ h)(x)[/tex] is
[tex]\left\{x\in\mathbb R \mid x\neq-1 \text{ and }x\neq0\right\}[/tex]
