The coefficient of $x^{-5}$ in the binomial expansion of ${\left(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right)}^{10}$, where $x\ne 0,1$ is

${\left(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right)}^{10}={\left[\frac{\left(x^{1/3}+1\right)\left(x^{2/3}-x^{1/3}+1\right)}{\left(x^{2/3}-x^{1/3}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]}^{10}$

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

General term in given expansion $={}^{10}{C}_r{\left(x^{1/3}\right)}^{10-r}{\left(\frac{-1}{x^{1/2}}\right)}^r={}^{10}{C}_r{x}^{\frac{10}{3}-\frac{r}{3}-\frac{r}{2}}{\left(-1\right)}^r$

Now for $x^{-5},\frac{10}{3}-\frac{r}{3}-\frac{r}{2}=-5$

$\Rightarrow \frac{10}{3}+5=r\left(\frac{1}{3}+\frac{1}{2}\right)\Rightarrow \frac{25}{3}=\frac{5}{6}r\Rightarrow r=\frac{25}{3}\times \frac{6}{5}=10$

$\therefore$ Coeff.of $x^{-5}={}^{10}{C}_{10}\left(1\right){\left(-1\right)}^{10}=1$