Answer
$f\circ g\circ h=\sqrt{\dfrac{\sqrt[3]{x}}{\sqrt[3]{x}-1}}$
Work Step by Step
$f(x)=\sqrt{x}$ $,$ $g(x)=\dfrac{x}{x-1}$ $,$ $h(x)=\sqrt[3]{x}$
First, find $g\circ h$ by substituting $x$ by $h(x)$ in $g(x)$ and simplifying if possible:
$g\circ h=g(h(x))=\dfrac{\sqrt[3]{x}}{\sqrt[3]{x}-1}$
Finally, to find $f\circ g\circ h$, substitute $x$ by $\dfrac{\sqrt[3]{x}}{\sqrt[3]{x}-1}$ in $f(x)$ and again, simplify if possible:
$f\circ g\circ h=f(g(h(x)))=\sqrt{\dfrac{\sqrt[3]{x}}{\sqrt[3]{x}-1}}$
