As the Equation 6.1.10 can be likely solved by an iterative numerical method (like Newton-Raphson) using the first derivative and successive point approximations. See this source for operational and convergence details.
The operational question is there a point in the realm of real numbers where the denominator of the first derivative goes to zero. The answer is no.
To demonstrate, if we did set the denominator equal to zero, take the square-root of both sides, and look for solutions indicating a possible singularity:
$ 1 + exp( -(x_i - \theta )) = 0 $
Or:
$ exp( -(x_i - \theta )) = -1 $
Or:
$ -(x_i - \theta ) = log(-1) $
To investigate a solution, I start with some background courtesy of Wikipedia, to quote:
For example, ln(i) = πi/2 or 5πi/2 or -3πi/2, etc.;
So, upon multiplying both sides of the first representation by two, I can successively develop an equivalence:
$2 \log(i) = log(i^2) = log(-1) = πi $
where my log function is more precisely defined as the natural log extended to the realm of complex numbers. Side note, I could have also used in place of πi/2 the value -πi/2, in which case the answer is also -πi.
Now, going back to our denominator set equal to zero, a singularity could theoretically arise when $\theta\ - x_i$ equals πi. However, this would imply either $\theta\ $ or $ x_i$ lie in the complex plane. which they do not.