By quantization, one is usually more trying to find the operator corresponding to a given symbol than the symbol corresponding to an operator. In your case, it seems the Weyl symbol you are talking about is the Wigner transform (the inverse of the Weyl trasnform). For an operator $A$ of kernel $a(x,y)$ it is defined by
$$
w_A(x,\xi) = ∫_{\mathbb{R}} e^{-i\,y\,\xi}\,a(x+y/2,x-y/2)\,\mathrm{d}y.
$$
If your operator is $A = -f(x)\,D^2_x = f(x)\,\partial_x^2$, then its kernel is $a(x,y) = f(x)\,\delta_0''(x-y)$, and so since $(x+y/2)-(x-y/2) = y$, we find (replacing the integral by a duality bracket)
$$
\begin{align*}
w_A(x,\xi) &= \left\langle \delta_0'', e^{-i\,\xi\,\cdot}f(x+\cdot/2)\right\rangle
\\
&= \left.\partial^2_y\left(e^{-i\,y\,\xi}f(x+y/2)\right)\right|_{y=0}
\\
&= -\xi^2 f(x) - i\,\xi\, f'(x) + \frac{1}{4}\, f''(x).
\end{align*}
$$