Question: $$x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}=\frac{\sqrt{(-\sqrt[3]{x^2}+1)}}{\sqrt[3]{x^4}}$$

Is it necessary to simplify the function further? What substitution is useful?

$u=\sqrt[n]{\frac{ax+b}{cx+d}}$ doesn't work.

Answer: 
As suggested in RecklessReckoner's comment, let us first change variable $x=u^3$ then $dx=3u^2du$. So, $$I=\int \frac{\sqrt{1-x^{2/3}}}{x^{4/3}}dx=3\int\frac{ \sqrt{1-u^2}}{u^2}du$$ Now, $u=\sin(t)$, $du=\cos(t)\,dt$ makes $$I=3\int \cot ^2(t)\,dt=3\int \frac{1-\sin^2(t)}{\sin^2(t)}dt =3\Big(\int \frac{dt}{\sin^2(t)}-\int dt \Big)=-3 \big(t+\cot (t)\big)$$ For sure, we could have saved a step with a single change of variable $x=\sin^3(t)$.
Following RecklessReckoner's comment:

We have $\int x^{\frac{-4}{3}}(-x^{\frac{2}{3}}+1)^{\frac{1}{2}}\mathrm dx$.

Now, let $u=x^{1/3}$ and $3u \mathrm du=\mathrm dx$

This gives $3\int u^{-4}(-u^{2}+1)^{\frac{1}{2}}u\mathrm du=3\int u^{-3}(-u^{2}+1)^{\frac{1}{2}}\mathrm du=3\int u^{-3}(1-u^{2})^{\frac{1}{2}}\mathrm du.$

Next, we perform another substitution: let $u=\sin t$ and $\mathrm du=\cos t \mathrm dt$

$3\int u^{-3}(1-u^{2})^{\frac{1}{2}}\mathrm du=3\int {\sin^{-3} t}(1-{\sin}^{2} t)^{\frac{1}{2}}\cos t\mathrm dt=3\int {\sin^{-3} t}({\cos}^{2} t)^{\frac{1}{2}}\cos t\mathrm dt$

$=3\int {\sin^{-3} t} \,{\cos}^{2} t\mathrm dt=3\int \frac{\cos^2t}{\sin^3 t} \mathrm dt=3\int {\cot^2t} \,{\csc t} \mathrm dt$

We can use the identity $\csc^2 t-1= \cot^2 t.$

$3\int {\cot^2t} \,{\csc t} \mathrm dt=3\int (\csc^2 t-1)\csc t \mathrm dt=3\int \csc^3 t-\csc t \mathrm dt=3\int \csc^3 t \mathrm dt- \int \csc t \mathrm dt.$

For the second term use the fact that $\int \csc t \mathrm dt =-\log|\csc t+\cot t| \mathrm dt+C$

For the first term, use the reduction formula $$\int \csc^m t \mathrm dt= \frac{(-\cos t)(\csc^{m-1} t)}{m-1}+\frac{m-2}{m-1}\int \csc^{m-2} t \mathrm dt$$

with m=3.

Then, substitute back for $t$ and $u$. Can you take it from here?