Using Taylor series expansion to solve the equation $\frac{\tanh^{-1}(x)}{\beta} -2x =0$

You should first see that, $$ \tanh^{-1} (x) = \frac {1}{2} \ln \left( \frac {1+x}{1-x} \right) $$ Then, you can use the second order Pade approximant, $$ \ln (1±x) ≈ \frac {±x(6±x)}{(6±4x)} $$ I hope it gives desired results.

Edit

The Pade approximant of the original function is, $$ \tanh^{-1} (x) ≈ \frac {x(15-4x^2)}{(15-9x^2)} $$ The solution to the resulting quadratic is, $$ x=± \sqrt \frac {15(2\beta -1)}{2(9\beta -2)}, \beta ≠ \frac {2}{9} $$ What remains to do is the limit, $$ \lim_{\beta → \frac {1}{2}} \sqrt {\frac {15}{9\beta -2}} = \sqrt {6} $$ I hope it helps.

Write the problem first as $$\beta=\frac{\tanh ^{-1}(x)}{2 x}$$ Now, expand the rhs using the usual $$\tanh ^{-1}(x)=\sum_{n=1}^\infty \frac {x^{2n+1}} {2n+1}$$ Using a few terms, we then have $$\beta=\frac{1}{2}+\frac{x^2}{6}+\frac{x^4}{10}+O\left(x^{6}\right)$$ Now, using series reversion $$\tilde{x} (\beta)=\sqrt{6} \sqrt{\beta -\frac{1}{2}}-\frac{9}{5} \sqrt{6} \left(\beta -\frac{1}{2}\right)^{3/2}+O\left(\left(\beta -\frac{1}{2}\right)^{5/2}\right)$$

as already given by ComplexYetTrivial.