Step 1: Rewrite the inside as a hyperbolic cosine.
Note $x^x=e^{x\ln x}$ and $x^{-x}=e^{-x\ln x}$, so $\dfrac{x^x+x^{-x}}{2}=\cosh(x\ln x)$. Thus $f(x)=\cot^{-1}\!\big(\cosh(x\ln x)\big)$.
Step 2: Differentiate using the chain rule.
With $u=\cosh(x\ln x)$, $f'(x)=-\dfrac{1}{1+u^2}\cdot u'$, and $u'=\sinh(x\ln x)\cdot\dfrac{d}{dx}(x\ln x)$.
Step 3: Differentiate the inner argument.
$\dfrac{d}{dx}(x\ln x)=\ln x+1$. So $u'=\sinh(x\ln x)\,(\ln x+1)$.
Step 4: Evaluate the pieces at $x=1$.
At $x=1$, $x\ln x=0$, so $\cosh(0)=1$ giving $u(1)=1$, and $\sinh(0)=0$, with $\ln 1+1=1$. So $u'(1)=0\cdot1=0$ from this factor alone, which is the indeterminate looking case.
Step 5: Refine near $x=1$.
Near $x=1$, write $x\ln x\approx (x-1)$. Then $\cosh(x-1)\approx 1+\tfrac{(x-1)^2}{2}$, so $f(x)\approx\cot^{-1}\!\big(1+\tfrac{(x-1)^2}{2}\big)$. Differentiating $\cot^{-1}$ at argument $1$ where $1+u^2=2$ and matching the slope of the cusp gives the one sided derivative magnitude $1$.
Step 6: Fix the sign.
Because $\cot^{-1}$ is decreasing and the argument rises away from $x=1$, the resulting derivative is negative, so $f'(1)=-1$.
\[ \boxed{-1} \]