1. Simplify the expression inside the bracket

We need the constant term in

$\left( \frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}} \right)^{10}.$

Let $t=x^{1/3}.$ Then $x=t^3$ and $x^{1/2}=t^{3/2}$ is not convenient, so instead simplify each fraction directly using algebraic identities.

2. First fraction

Use $a^2-a+1=\frac{a^3+1}{a+1}$ with $a=x^{1/3}$. Since $x^{2/3}-x^{1/3}+1=t^2-t+1=\frac{t^3+1}{t+1}=\frac{x+1}{x^{1/3}+1},$ we get

$\frac{x+1}{x^{2/3}-x^{1/3}+1}=x^{1/3}+1.$

3. Second fraction

Factor the denominator: $x-x^{1/2}=x^{1/2}(x^{1/2}-1).$ Also, $x-1=(x^{1/2}-1)(x^{1/2}+1).$ Hence

$\frac{x-1}{x-x^{1/2}}=\frac{(x^{1/2}-1)(x^{1/2}+1)}{x^{1/2}(x^{1/2}-1)}=\frac{x^{1/2}+1}{x^{1/2}}=1+x^{-1/2}.$

4. Expression inside the bracket

Therefore, $\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}=(x^{1/3}+1)-(1+x^{-1/2})=x^{1/3}-x^{-1/2}.$

So the given expression becomes

$\left(x^{1/3}-x^{-1/2}\right)^{10}.$

5. General term

In the binomial expansion, $T_{r+1}=\binom{10}{r}(x^{1/3})^{10-r}(-x^{-1/2})^r.$

Thus the power of $x$ in the general term is

$\frac{10-r}{3}-\frac{r}{2}.$

For the term independent of $x$, set this exponent to $0$:

$\frac{10-r}{3}-\frac{r}{2}=0.$

Multiply by $6$: $2(10-r)-3r=0$ $20-2r-3r=0$ $20-5r=0$ $r=4.$

6. Constant term

For $r=4$, $T_5=\binom{10}{4}(x^{1/3})^6(-x^{-1/2})^4.$ Since the power of $x$ is zero, the coefficient is

$\binom{10}{4}(-1)^4=\binom{10}{4}=210.$

7. Final answer

The term independent of $x$ is

$\boxed{210}.$