Double Differentiation or Double Trouble?

First note since $\alpha$ is a root of $f(x)$ ,

$\alpha^4-3\alpha^3-6\alpha^2+9\alpha+6=0$

$\implies \color{#0C6AC7}{\alpha^4-3\alpha^3=6\alpha^2-9\alpha-6~...(1)}$

Now double differentiate $f(x)$ and put $x=\alpha$ so that $f''(\alpha)=12\alpha^2-18\alpha-12=2 (\color{#0C6AC7}{6\alpha^2-9\alpha-6})$ . Now using $(1)$ , $f''(\alpha)=2(\color{#0C6AC7}{\alpha^4-3\alpha^3})=2\alpha^3(\alpha -3)$ .

Hence,

$\prod_{\text{cyc}}f''(\alpha)=\displaystyle\prod_{\text{cyc}}2\alpha^3(\alpha-3)=2^4(\alpha\beta\gamma\delta)^3\displaystyle\prod_{\text{cyc}}(\alpha-3)$

Using Vieta's formula $\small{\alpha\beta\gamma\delta=6}$ and put $x=3$ in $f(x)=\displaystyle\prod_{\text{cyc}}(\alpha-x)$ to get $\displaystyle\prod_{\text{cyc}}(\alpha -3)=f(3)$ . Hence,

$\displaystyle\prod_{\text{cyc}}f''(\alpha)=2^4\cdot (6)^3\cdot f(3)$

$\implies \large \dfrac{\displaystyle\prod_{\text{cyc}}f''(\alpha)}{2^3\cdot 6^2\cdot f(3)}=\boxed{12}$

$f(x)=x^4-3x^3-6x^2+9x+6=\prod (x-\alpha)=\prod (\alpha-x) \quad \quad \small{\text{(1)}}$

$f''(x)=12x^2-18x-12=12 \left (x+\dfrac{1}{2} \right) \left (x-2 \right)$

$\prod f''(\alpha)=\prod 12 \left (\alpha+\dfrac{1}{2} \right) \left (x-2 \right)=12^4 \left (\prod \left (\alpha+\dfrac{1}{2} \right) \right) \left (\prod \left (\alpha-2 \right) \right)$

$\cdots=12^4 f \left (-\dfrac{1}{2} \right) f \left (2 \right)=12^4 \left (\dfrac{7}{16} \right) \left (-8 \right)=-72576 \quad \quad \small{\color{#3D99F6}{\text{Using the result for f(x) from (1)}}}$

$\mathcal Y =\dfrac{\left (\prod f''(\alpha) \right)}{2^3 \cdot 6^2 \cdot f(3)}=\dfrac{-72576}{288 \times (-21)}=\boxed{\boxed{12}}$