Given $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$

Let $t=\sqrt{\frac{x}{1-x}}$

i.e. $\frac{x}{1-x}>0 \& x\ne 0,1$

i.e. $x\left(1-x\right)>0\Rightarrow x\left(x-1\right)<0\Rightarrow x\in \left(0,1\right)$

Now, $t+\frac{1}{t}=\frac{13}{6}\Rightarrow 6t^2-13t+6=0$

$\Rightarrow t=\frac{13\pm \sqrt{169-144}}{12}=\frac{3}{2}, \frac{2}{3}$

i.e. when $\sqrt{\frac{x}{1-x}}=\frac{3}{2}\Rightarrow \frac{x}{1-x}=\frac{9}{4}$

$\Rightarrow 4x=9-9x\Rightarrow x=\frac{9}{13}$

when $\sqrt{\frac{x}{1-x}}=\frac{2}{3}\Rightarrow \frac{x}{1-x}=\frac{4}{9}$

$\Rightarrow 9x=4-4x\Rightarrow x=\frac{4}{13}$

Hence, there are two real roots.