If $\int\frac{\sqrt{x}dx}{\sqrt{1-x^2}}=\frac{2}{3}fog(x)+c$ then:

$g(x)=x\sqrt{x}$

Let $1-x^3=t^2⇒\int\frac{\sqrt{x}dx}{\sqrt{1-x^2}}=\int\frac{-2dt}{3x^{3/2}}=\int\frac{-2}{3}\frac{dt}{\sqrt{1-t^2}}=\frac{2}{3}\cos^{-1}(t)=\frac{2}{3}\cos^{-1}(\sqrt{1-x^3})$

But $\cos^{-1}\sqrt{1-x^3}$ can also be written as $\sin^{-1}(x^{3/2})$

$⇒f(x)=\sin^{-1}x$ and $g(x) = x^{3/2}=x\sqrt{x}$