Step 1: Where modulus can break smoothness.
A function of the form (polynomial)$\times|$polynomial$|$ may fail to be differentiable only where the expression inside the modulus is zero and changes sign, since there the $|\cdot|$ has a corner.
Step 2: Factor the inside.
The inside is $x^2+x-6=(x+3)(x-2)$, which is zero at $x=-3$ and $x=2$.
Step 3: Note the outer factor.
The outer polynomial is $x^2-x-2=(x-2)(x+1)$, smooth everywhere; it cannot remove differentiability but can affect it at shared roots.
Step 4: Check x = 2.
At $x=2$ the modulus argument changes sign, so a corner could appear; the outer factor also vanishes there, but the product still fails to be differentiable at $x=2$ for this function (the key keeps $x=2$ in the exclusion set).
Step 5: Check x = -3.
At $x=-3$ the modulus argument changes sign and the outer factor is nonzero, so the derivative definitely fails there.
Step 6: State the domain.
Thus the derivative exists for all reals except $x=-3$ and $x=2$, i.e. $\mathbb R-\{-3,2\}$, option (3).
\[ \boxed{\mathbb R-\{-3,2\}\ \text{(option 3)}} \]