The given equation is

$\begin{array}{l}
\frac{1}{x}-\frac{1}{x-2}=3, x \neq 0,2 \\
\Rightarrow \frac{x-2-x}{x(x-2)}=3 \\
\Rightarrow \frac{-2}{x^{2}-2 x}=3 \\
\Rightarrow-2=3 x^{2}-6 x \\
\Rightarrow 3 x^{2}-6 x+2=0
\end{array}$

This equation is of the form $a x^{2}+b x+c=0$, where $a=3, b=-6$ and $c=2$.

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-6)^{2}-4 \times 3 \times 2=36-24=12>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{12}=2 \sqrt{3}$

$\begin{array}{l}
\therefore a=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-6)+2 \sqrt{3}}{2 \times 3}=\frac{6+2 \sqrt{3}}{6}=\frac{3+\sqrt{3}}{3} \\
\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-6)-2 \sqrt{3}}{2 \times 3}=\frac{6-2 \sqrt{3}}{6}=\frac{3-\sqrt{3}}{3}
\end{array}$

Hence, $\frac{3+\sqrt{3}}{3}$ and $\frac{3-\sqrt{3}}{3}$ are the roots of the given equation.