Let $d\in\mathbb N$ and $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$. If $f\in\mathcal L^1(\lambda^{\otimes d})$, then $$\hat f(y):=\frac1{(2\pi)^d}\int e^{{\rm i}\langle x,\:y\rangle}f(x)\:\lambda^{\otimes d}({\rm d}x)\;\;\;\text{for }y\in\mathbb R^d.$$
How can we show that if $f\in C^2_c(\mathbb R^d)$, then $\hat f\in\mathcal L^1(\lambda^{\otimes d})$?
The claim is made in this post (for the case $d=1$).
I'm not sure if this is the right approach, but maybe we need to apply a Taylor expansion: Let $$e_y(x):=e^{{\rm i}\langle x,\:y\rangle}\;\;\;\text{for }x\in\mathbb R^d$$ for $y\in\mathbb R^d$. Then, for every fixed $x\in\mathbb R^d$, there is a $\xi\in B_{\|x\|}(0)$ with $$e_y(x)=1+{\rm i}\langle x,y\rangle-\frac12\left|\langle x,y\rangle\right|^2\underbrace{-{\rm i}\frac16e_y(\xi)\langle x,y\rangle^3}_{=:\:R(x)}\tag2$$ and $$\left|R(x)\right|\le\left|\langle x,y\rangle\right|^3\tag3.$$
However, don't see how we can utilize this ...
If $f \in C^2_c(\mathbb R)$, then integrate by parts in the definition of the Fourier transform to get, for some constant $C$ which may change from line to line, \begin{align*} |(1+\xi^2)\widehat f(\xi)| &\le C|\widehat f(\xi) + \widehat{f''}(\xi)|\le C(\|f\|_1 + \|f''\|_1). \end{align*} Thus, \begin{align*} |\widehat f(\xi)| \le \frac{C}{1+\xi^2}, \end{align*} which is integrable on $\mathbb R$.
For arbitrary $d\ge 1$, an analogous argument using all the second order partials of $f$ would give a bound of the form \begin{align*} |\widehat f(\xi_1,\dots,\xi_d)| \le \frac{C}{1+|(\xi_1,\dots,\xi_d)|^2}, \end{align*} which is only integrable if $d=1$. I don't have a concrete counterexample for $d\ge 2$, but I suspect you might be able to make one by playing with functions that are $C^2_c$ and not $C^3_c$ like in this answer.