$\displaystyle{y=\frac{x \, \sqrt[]{x^2+1}}{(x+1)^{2/3}}}$
Let $f=x \, \sqrt[]{x^2+1}$ and $g=(x+1)^{2/3}$.
Then $\displaystyle{y'=\frac{f'\cdot g-f\cdot g'}{g^2}} \ \ \ (\star)$
We have the following:
\begin{align*}f' & =(x)' \, \sqrt[]{x^2+1}+x \, (\sqrt[]{x^2+1})'= \sqrt[]{x^2+1}+x \, \frac{1}{2\sqrt[]{x^2+1}}\cdot (x^2+1)' =\sqrt[]{x^2+1}+x \, \frac{2x}{2\sqrt[]{x^2+1}} \\ &=\sqrt[]{x^2+1}+ \, \frac{x^2}{\sqrt[]{x^2+1}} =\frac{\sqrt[]{x^2+1}^2+x^2}{\sqrt[]{x^2+1}}=\frac{x^2+1+x^2}{\sqrt[]{x^2+1}}\\ &=\frac{2x^2+1}{\sqrt[]{x^2+1}}\end{align*}
$$g'=\frac{2}{3}(x+1)^{2/3-1}=\frac{2}{3}(x+1)^{-1/3}$$
$$g^2=(x+1)^{4/3}$$
So, substituting these at the relation $(\star)$ we get:
\begin{align*}y' &=\frac{\frac{2x^2+1}{\sqrt[]{x^2+1}}\cdot (x+1)^{2/3}-x \, \sqrt[]{x^2+1}\cdot \frac{2}{3}(x+1)^{-1/3}}{(x+1)^{4/3}} \\ &=\frac{\sqrt{x^2+1}\left (\frac{2x^2+1}{\sqrt[]{x^2+1}}\cdot (x+1)^{2/3}-x \, \sqrt[]{x^2+1}\cdot \frac{2}{3}(x+1)^{-1/3}\right )}{(x+1)^{4/3}\sqrt{x^2+1}} \\ &=\frac{\left (2x^2+1\right )\cdot (x+1)^{2/3}-x \, (x^2+1)\cdot \frac{2}{3}(x+1)^{-1/3}}{(x+1)^{4/3}\sqrt{x^2+1}} \\ &=\frac{(x+1)^{1/3}\left (\left (2x^2+1\right )\cdot (x+1)^{2/3}-x \, (x^2+1)\cdot \frac{2}{3}(x+1)^{-1/3}\right )}{(x+1)^{1/3}(x+1)^{4/3}\sqrt{x^2+1}} \\ &=\frac{\left (2x^2+1\right )\cdot (x+1)-x \, (x^2+1)\cdot \frac{2}{3}}{(x+1)^{5/3}\sqrt{x^2+1}} \\ & =\frac{3\left (\left (2x^2+1\right )\cdot (x+1)-x \, (x^2+1)\cdot \frac{2}{3}\right )}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{3\left (2x^2+1\right )\cdot (x+1)-2x \, (x^2+1)}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{3\left (2x^3+x+2x^2+1\right )-2 \, (x^3+x)}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{6x^3+3x+6x^2+3-2x^3-2x}{3(x+1)^{5/3}\sqrt{x^2+1}} \\ &=\frac{4x^3+6x^2+x+3}{3(x+1)^{5/3}\sqrt{x^2+1}}\end{align*}