Step 1: Write the derivative on each open interval.
For \(-1 \lt x \lt 1\), \(f(x)=\tan^{-1}x\) so \(f'(x)=\frac{1}{1+x^2}\). For \(x \gt 1\), \(f(x)=\frac{x-1}{2}\) so \(f'(x)=\frac{1}{2}\). For \(x \lt -1\), \(f(x)=\frac{-x-1}{2}\) so \(f'(x)=-\frac{1}{2}\).
Step 2: Compare one-sided derivatives at x = 1.
Left-hand derivative: \(\lim_{x \to 1^-}\frac{1}{1+x^2}=\frac{1}{2}\). Right-hand derivative: \(\frac{1}{2}\) (constant piece). These agree, so \(f'(1)\) exists and equals \(\frac{1}{2}\).
Step 3: Compare one-sided derivatives at x = -1.
Left-hand derivative: \(-\frac{1}{2}\) (constant piece). Right-hand derivative: \(\lim_{x \to -1^+}\frac{1}{1+x^2}=\frac{1}{2}\). These do not agree, so \(f'(-1)\) does not exist.
Step 4: Final conclusion.
The derivative exists at every real number except at \(x=-1\), so the domain of \(f'(x)\) is
\[
\boxed{\mathbb{R}-\{-1\}}
\]