Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

So we'll take your textbook's definition: $$ \mbox{arccot}(x)=\frac{\pi}{2}-\arctan(x). $$

Then $$ \lim_{0^+} \mbox{arccot} (x)=\frac{\pi}{2}=\lim_{0^+} \arctan(1/x) $$ while $$ \lim_{0^-} \mbox{arccot} (x)=\frac{\pi}{2}=-\lim_{0^-} \arctan(1/x). $$ Since its derivative is $0$ on $\mathbb{R}^*$, $$ \mbox{arccot}(x)-\arctan(1/x)=0 $$ for all $x>0$ while $$ \mbox{arccot}(x)-\arctan(1/x)=\pi $$ for all $x<0$.

This does not contradict the zero-derivative theorem because the function is not defined at $0$, so its domain is not connected.

This is not a rigorous proof, it is intuitive one.

Let $y = \arctan\left(\dfrac1x\right)$ then $$ \begin{align} \tan(y) &= \frac1x\\ \frac1{\tan(y)} &= x\\ \cot(y) &= x\\ y &= \text{arccot}(x)\\ \arctan\left(\frac1x\right) &= \text{arccot}(x)\\ \arctan\left(\frac1x\right) - \text{arccot}(x)&=0. \end{align} $$

It holds for $x \neq 0, \ x\in\mathbb{R}$.