But differentiating three times might be too long...

$\begin{aligned} y & = \frac {3x^2+12x+11}{x^3+6x^2+11x+6} = \frac {3(x+2)^2 - 1}{(x+1)(x+2)(x+3} & \small \blue{\text{Let } u = x+2 \implies \frac {du}{dx} = 1} \\ & = \frac {3u^2-1}{(u-1)u(u+1)} = \frac {3(u^2-1)+2}{u(u^2-1)} \\ & = \frac 3u + \frac 2{(u-1)u(u+1)} = \frac 3u + \frac 1{u-1} - \frac 2u + \frac 1{u+1} \\ & = \frac 1{u-1} + \frac 1u + \frac 1{u+1} \\ \frac {dy}{dx} & = \frac {dy}{du}\cdot \cancel{\dfrac {du}{dx}}^1 = - \frac 1{(u-1)^2} - \frac 1{u^2} - \frac 1{(u+1)^2} \\ \frac {d^2y}{dx^2} & = \frac d{du} \left(\frac {dy}{dx}\right)\cancel{\dfrac {du}{dx}}^1 = \frac 2{(u-1)^3} + \frac 2{u^3} + \frac 2{(u+1)^3} \\ \frac {d^3y}{dx^3} & = \frac d{du} \left(\frac {d^2y}{dx^2}\right)\cancel{\dfrac {du}{dx}}^1 = -\frac 6{(u-1)^4} - \frac 6{u^4} - \frac 6{(u+1)^4} \\ \frac {d^3y}{dx^3}\bigg|_{x=0} & = \frac {d^3y}{du^3}\bigg|_{u=2} = -\frac 6{1^4} - \frac 6{2^4} - \frac 6{3^4} = - \frac {1393}{216} \end{aligned}$

Therefore, $a+b = 1393+216 = \boxed{1609}$ .

I think we can make the computation slightly easier by recognizing that we can set y(x) = f'(x) / f(x), where y(x) is the function being differentiated and f(x) = x^3 + 6 x^2 + 11 x + 6. Then,

y(x)*f(x) = f'(x).

Now, apply the product rule 3 times and note that f(0) = 6, f'(0) = 11, f''(0) = 12, f'''(0) = 6, and all higher-order derivatives vanish. It's still a long computation, though.

We have $x^3 + 6x^2 + 11x + 6 = (x + 1)(x + 2)(x + 3)$ so we see that it is easier to solve this problem if we do partial fraction decomposition first. We have $\frac{3x^2 + 12x + 11}{x^3 + 6x^2 + 11x + 6} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$ or, multiplying both sides by $(x + 1)(x + 2)(x + 3)$ , $3x^2 + 12x + 11 = A(x + 2)(x + 3) + B(x + 1)(x + 3) + C(x + 1)(x + 2)$ If $x=-1$ , the $B$ and $C$ terms vanish, leaving us with $3(-1)^2 + 12(-1)+12=A(-1+2)(-1+3)+0+0\Longrightarrow A = 1$ If $x=-2$ , the $A$ and $C$ terms zero out, so that $3(-2)^2 + 12(-2)+12=0+B(-2+1)(-2+3)+0\Longrightarrow B =1$ If $x=-3$ , the $A$ and $B$ terms disappear, hence $3(-3)^2 + 12(-3)+12=0+0+C(-3+1)(-3+2)\Longrightarrow C =1$ Therefore, $f(x) = \frac{3x^2 + 12x + 11}{x^3 + 6x^2 + 11x + 6} = \frac{1}{x + 1} + \frac{1}{x + 2} + \frac{1}{x + 3}$ Now, each fraction is way easier to differentiate than the original one. We proceed as $\begin{aligned} f'(x) &= -\frac{1}{(x+1)^2}-\frac{1}{(x+2)^2}-\frac{1}{(x+3)^2} \\ f''(x) &= \frac{2}{(x+1)^3}+\frac{2}{(x+2)^3}+\frac{2}{(x+3)^3} \\ f'''(x) &= -\frac{6}{(x+1)^4}-\frac{6}{(x+2)^4}-\frac{6}{(x+3)^4} \\ \therefore f'''(0) &= -\frac{6}{(0+1)^4}-\frac{6}{(0+2)^4}-\frac{6}{(0+3)^4}=-\frac{1393}{216} \\ \end{aligned}$ giving $a+b=1393+216=\boxed{1609}$ .