Answer
$$\tanh^{-1}\Big(-\frac{1}{2}\Big)=-\frac{\ln3}{2}$$
Work Step by Step
We have $$\tanh^{-1}x=\frac{1}{2}\ln\frac{1+x}{1-x}$$ for $|x|\lt1$
Therefore, $$\tanh^{-1}\Big(-\frac{1}{2}\Big)=\frac{1}{2}\ln\frac{1-1/2}{1+1/2}$$ $$=\frac{1}{2}\ln (\frac{1}{2}\div\frac{3}{2})$$ $$=\frac{1}{2}\ln(\frac{1}{3})$$ $$=-\frac{\ln3}{2}$$
