Step 1: Plan:
Compute $y'(0)$ directly from the sum after seeing that every term has the same pattern.
Step 2: Pattern:
The $k$th term is $\tan^{-1}\frac{1}{x^2+(2k-1)x+k^2-k+1}$, which equals $\tan^{-1}(x+k) - \tan^{-1}(x+k-1)$. Adding $k=1$ to $n$, all middle terms cancel, leaving $\tan^{-1}(x+n)-\tan^{-1}x$.
Step 3: Slope at zero:
Derivative of $\tan^{-1}u$ is $\frac{u'}{1+u^2}$. At $x = 0$ we get $\frac{1}{1+n^2}-\frac{1}{1} = \frac{1-(1+n^2)}{1+n^2} = -\frac{n^2}{n^2+1}$.
This is negative, so options A and D (positive) and C (zero, impossible for $n\ge1$) are wrong.
Final Answer:
$y'(0) = -\dfrac{n^2}{n^2+1}$, option (B).
\[ \boxed{-\frac{n^2}{n^2+1}} \]