Step 1: Split the integrand using odd-even analysis.
$I = \int_{-1}^{1} \dfrac{\sin x - x^2}{3-|x|}\,dx$. Since $3-|x|$ is even and $\sin x$ is odd, $\dfrac{\sin x}{3-|x|}$ is odd, so its integral over $[-1,1]$ is 0.
Step 2: Reduce to the even part.
\[ I = -\int_{-1}^{1} \frac{x^2}{3-|x|}\,dx = -2\int_0^1 \frac{x^2}{3-x}\,dx. \] (Using even function property.)
Step 3: Perform polynomial long division.
$\dfrac{x^2}{3-x}$: Since $x^2 = -(3-x)\cdot(-x-3) + 9 - ...$, try: $x^2 = (3-x)(-x-3) + 9$. Check: $(3-x)(-x-3) = -3x - 9 + x^2 + 3x = x^2 - 9$. So $x^2 = (3-x)(-x-3) + 9$. Hmm, this gives a remainder. Let us do it properly: divide $x^2$ by $(3-x) = -(x-3)$. $\dfrac{x^2}{-(x-3)} = -x - 3 - \dfrac{9}{x-3}$. So $\dfrac{x^2}{3-x} = -x - 3 + \dfrac{9}{3-x}$... Check: $(-x-3)(3-x) = -3x+x^2-9+3x = x^2-9$, so $x^2 = (-x-3)(3-x)+9$. Thus $\dfrac{x^2}{3-x} = -x-3 + \dfrac{9}{3-x}$.
Step 4: Integrate.
\[ \int_0^1 \frac{x^2}{3-x}\,dx = \int_0^1\left(-x-3+\frac{9}{3-x}\right)\,dx = \left[-\frac{x^2}{2}-3x-9\ln(3-x)\right]_0^1. \] $= \left(-\frac{1}{2}-3-9\ln 2\right) - \left(0-0-9\ln 3\right) = -\frac{7}{2} - 9\ln 2 + 9\ln 3 = -\frac{7}{2} + 9\ln\frac{3}{2}$.
Step 5: Multiply by $-2$.
$I = -2\left(-\dfrac{7}{2}+9\ln\dfrac{3}{2}\right) = 7 - 18\ln\dfrac{3}{2} = 7 - 18\log\dfrac{3}{2}$.
Step 6: State the answer.
$I = 7 - 18\log\dfrac{3}{2}$.
\[ \boxed{7 - 18\log\dfrac{3}{2}} \]