First notice that

$\displaystyle \displaystyle \alpha=\tan^{-1}\left( \sqrt[3]{\frac{y}{x}}\right) \iff tan(\alpha)=\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}} }$

Since $\displaystyle \displaystyle 0 < \alpha \le \frac{\pi}{2}$ it must be in the first quadrant. Use the Pythagorean theorem to find the hypotenuse of the triangle. see diagram

Using this we get that

$\displaystyle \displaystyle \sin(\alpha)=\frac{y^{\frac{1}{3}}}{\sqrt{x^{\frac {2}{3}}+y^{\frac{2}{3}}}} \text{ and } \cos(\alpha)=\frac{x^{\frac{1}{3}}}{\sqrt{x^{\frac {2}{3}}+y^{\frac{2}{3}}}}$

Now just plug this in to get

$\displaystyle \displaystyle M(\alpha)=\frac{y}{\frac{y^{\frac{1}{3}}}{\sqrt{x^ {\frac{2}{3}}+y^{\frac{2}{3}}}}}+\frac{x}{\frac{x^ {\frac{1}{3}}}{\sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3 }}}}}=y^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^{\fra c{2}{3}}}}+x^{\frac{2}{3}}\sqrt{x^{\frac{2}{3}}+y^ {\frac{2}{3}}}}=$

$\displaystyle \displaystyle \sqrt{x^{\frac{2}{3}}+y^{\frac{2}{3}}}}\left( x^{\frac{2}{3}}+y^{\frac{2}{3}}}\right)=\left(x^{\ frac{2}{3}}+y^{\frac{2}{3}}}\right)^{\frac{3}{2}}$