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: 
Notice, $$\int x^{-4/3}\left(-x^{2/3}+1\right)^{1/2}\ dx$$$$=\int \frac{1}{x}\left(-x^{2/3}+1\right)^{1/2}(x^{-1/3}\ dx)$$

Let $-x^{2/3}+1=\sin^2\theta\implies -\frac{2}{3}x^{-1/3}\ dx=2\sin\theta\cos\theta\ d\theta$ or $x^{-1/3}\ dx=-3\sin\theta\cos\theta\ d\theta $
$$=\int\frac{1}{(1-\sin^2\theta)^{3/2}}(\sin\theta)(-3\sin\theta\cos\theta\ d\theta )$$
$$=-3\int\frac{\sin^2\theta\cos\theta}{\cos^3\theta}\ d\theta$$
$$=-3\int\tan^2\theta\ d\theta$$
$$=-3\int(\sec^2\theta-1)\ d\theta$$

$$=-3(\tan\theta-\theta)+C$$
$$=3\theta-3\tan\theta+C$$
substituting back the value of $\theta$, 
$$=\color{red}{3\sin^{-1}(\sqrt{1-x^{2/3}}) \ - \ 3 \frac{\sqrt{x^{-2/3}-1}}{x^{1/3}} \ + \ C}$$
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?