Problem by Lee Dongheng

Interchange $y\to-y$ and the equations remain same hence for each $y$ there also exist $-y$ which is also a solution of the given equations. Hence sum $=\boxed{0}$ .

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$\Large\mathcal{\color{#0C6AC7}{Alternate~Solution}}$

Substitute $\sqrt x=p,\sqrt[3]{y^2}=q$ $\large \begin {cases} p+q=8\\p^2+q^2=32\end {cases}$ Substituting $p=8-q$ in the second equation and solving:- $q^2-8q+16=0\implies (q-4)^2=0\implies q=4$

$\implies y=(4)^{3/2}=\pm 8$ Hence sum $=8-8=\boxed{0}~~,$ which is exactly what we conclude earlier.

Anyways, $x=4$ .

$\large \begin {cases} x^\frac{1}{2}+y^\frac{2}{3}=8 ...(1) \\x+y^\frac{4}{3}=32 ...(2)\end {cases}$

Squaring first equation.

$\underbrace {x+y^{4/3}}_{32}+2×x^{1/2}×y^{2/3}=64$

$\underbrace {x^{1/2}}_{8-y^{2/3}}×y^{2/3}=16$

$8y^{4/3}-8y^{2/3}-16=0$

Use Quadratic formula to get,

$y^{2/3}=4$

$\therefore y=\pm 8$

Hence, sum of all values of $y=8-8=\boxed{0}$