Step 1: Recognize the substitution that simplifies the equation.
The equation $x^{2/3}+x^{1/3}-2=0$ has the form of a quadratic in $x^{1/3}$. This is the key observation: if we let $t = x^{1/3}$, then $x^{2/3} = (x^{1/3})^2 = t^2$, and the equation becomes $t^2+t-2=0$, which is a standard quadratic.
Step 2: Solve the quadratic in $t$.
$t^2+t-2=0$. We factor: look for two numbers multiplying to $-2$ and adding to $+1$. Those are $+2$ and $-1$. So $(t+2)(t-1)=0$, giving $t=-2$ or $t=1$.
Step 3: Convert back from $t$ to $x$.
Since $t = x^{1/3}$, we cube to get $x = t^3$. For $t=1$: $x=1^3=1$. For $t=-2$: $x=(-2)^3=-8$. The two roots of the original equation are $x=1$ and $x=-8$.
Step 4: Verify both roots satisfy the original equation.
For $x=1$: $1^{2/3}+1^{1/3}-2 = 1+1-2 = 0$. Correct. For $x=-8$: $(-8)^{2/3}+(-8)^{1/3}-2 = 4+(-2)-2 = 0$. Correct.
Step 5: Compute the sum of squares of the roots.
Sum of squares $= 1^2+(-8)^2 = 1+64 = 65$.
Step 6: State the final answer.
The sum of squares of the roots is $65$. \[ \boxed{65} \]