Step 1: Understand the goal.
We have $y$ as a sum of two inverse tangent terms and we must find $\frac{dy}{dx}$. Differentiating directly is messy, so first we simplify $y$.
Step 2: Recall the splitting rule.
A useful identity is \[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right). \] We will write each given term as such a difference, choosing $A$ and $B$ that differ by $1$.
Step 3: Rewrite the first term.
Take $A = x+1$ and $B = x$. Then $A - B = 1$ and $1 + AB = 1 + x(x+1) = 1 + x + x^2$. So \[ \tan^{-1}\left(\frac{1}{1+x+x^2}\right) = \tan^{-1}(x+1) - \tan^{-1}(x). \]
Step 4: Rewrite the second term.
Take $A = x+2$ and $B = x+1$. Then $A - B = 1$ and $1 + AB = 1 + (x+1)(x+2) = x^2 + 3x + 3$. So \[ \tan^{-1}\left(\frac{1}{x^2+3x+3}\right) = \tan^{-1}(x+2) - \tan^{-1}(x+1). \]
Step 5: Add and cancel.
Adding the two parts, the $\tan^{-1}(x+1)$ terms cancel: \[ y = \tan^{-1}(x+2) - \tan^{-1}(x). \]
Step 6: Differentiate the clean form.
Using $\frac{d}{dx}\tan^{-1}(u) = \frac{u'}{1+u^2}$ with $u' = 1$ for both terms, \[ \frac{dy}{dx} = \frac{1}{1+(x+2)^2} - \frac{1}{1+x^2}. \] This is the same as $-\frac{1}{1+x^2} + \frac{1}{1+(x+2)^2}$. \[ \boxed{\dfrac{dy}{dx} = \dfrac{-1}{1+x^2} + \dfrac{1}{1+(x+2)^2}} \]