Answer
See below.
Work Step by Step
If the derivative of $f(x)$ is zero, then f is a constant function.
Let $f(x)=\displaystyle \tan^{-1}x+\tan^{-1}(\frac{1}{x})$
Use $\displaystyle \frac{d}{dx}(\tan^{-1}u)=\frac{1}{1+u^{2}}\cdot\frac{du}{dx}$,
$\displaystyle \frac{d}{dx}[f(x)]=\frac{1}{1+x^{2}}+\frac{1}{1+(\frac{1}{x})^{2}}\cdot(-x^{-2})$
$=\displaystyle \frac{1}{1+x^{2}}-\frac{1}{x^{2}(1+\frac{1}{x^{2}})}$
$=\displaystyle \frac{1}{1+x^{2}}-\frac{1}{x^{2}+1}=0$
So, $f(x)=\displaystyle \tan^{-1}x+\tan^{-1}(\frac{1}{x})=C$
which is a constant function.
