solve this

Let $x=\sin ^{\frac{2}{3}} t$

Differentiate both side with respect to $t$

$\frac{d x}{d t}=\frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t \Rightarrow d x=\frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t d t$

$\frac{d x}{d t}=\frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t \Rightarrow d x=\frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t d t$

$y=\int \frac{1}{\sin ^{\frac{2}{3}} t \sqrt{1+\sin ^{2} t}} \frac{2}{3} \sin ^{-\frac{1}{3}} t \cos t d t$

$y=\frac{2}{3} \int \operatorname{cosec} t d t$

$y=\frac{2}{3} \ln (\operatorname{cosec} t-\cot t)+c$

Again, put $t=\sin ^{-1} \chi^{\frac{3}{2}}$

$y=\frac{2}{3} \ln \left(\operatorname{cosec} \sin ^{-1} x^{\frac{3}{2}}-\cot \sin ^{-1} x^{\frac{3}{2}}\right)+c$

$y=\frac{2}{3} \ln \left(x^{\frac{-3}{2}}-\frac{\sqrt{1-x^{3}}}{x^{\frac{3}{2}}}\right)+c$

$y=-\ln x+\frac{2}{3} \ln \left(1-\sqrt{1-x^{3}}\right)+c$