Suppose that $f\in \mathcal{S}(\mathbb R^{2})$ (Schwartz space). Let $\bar{y}=(y_{1}, y_{2}) \in \mathbb R^{2}.$ Define $F:\mathbb R^{2}\to \mathbb C$ as follows: $$F(\bar{x})= f(\bar{x}-\bar{y})e^{i(x_{1}y_2- x_{2}y_{1})}, \ \ \bar{x}=(x_1, x_2)\in \mathbb R^2$$
My Question is: Can we expect $F\in \mathcal{S}(\mathbb R^{2})$?
(My vague thought: translation of a Schwartz function is Schwartz function but I don't know what happen if we multiply with modulating factor)
Assume $g \in \mathcal{S}$. The derivatives $\partial_{x_1}^i \partial_{x_2}^j e^{i (x_1 y_2 - y_2 x_1)}$ are of the form $P_{i,j}(x) e^{i( x_1 y_2 - y_2 x_1)}$ where $P_{i,j}$ is some polynomial (actually, just constant). Thus $g(x_1,x_2) e^{i (x_1 y_2 - y_2 x_1)}$ is still a Schwartz function (since the absolute value of the exponential term is $1$).
Edit: In more detail (I'll work in one dimension for clarity): Since Schwartz functions, by definition, are smooth functions $f$ such that $|f(x)|$ and and its derivatives $|f^{(n)}(x)|$ decrease faster than any polynomial as $|x| \to \infty$, it makes sense that $e^{ix}$ times a Schwartz function is still a Schwartz function.
Formally, Schwartz functions must satisfy $$ \sup_{x \in \Bbb{R}} |x^n D^m f(x)| < \infty $$ for each $n,m \in \Bbb{N}$ and $D^m$ denotes the $m$th derivative. This is just a restatement of "f and its derivatives decrease faster than polynomials". Now plug in $f(x) e^{ix}$, and we have $$ \sup_{x \in \Bbb{R}} |x^n D^m(f(x) e^{ix})|\,. $$ First, assume $m=0$, and this is just $|\sup |x^n f(x) e^{ix}| = \sup |x^n f(x)|$ which is finite because $f \in \mathcal{S}$.
Now, if $m = 1$ we get \begin{align} \sup_{x \in \Bbb{R}} |x^n D^m(f(x) e^{ix})| &= \sup_{x \in \Bbb{R}} |x^n (f'(x) e^{ix} + f(x) i e^{ix})| \\ & \leq \sup_{x \in \Bbb{R}} |x^n f'(x)| + \sup_{x \in \Bbb{R}} |x^n f(x)|\,, \end{align} which is again finite since $f \in \mathcal{S}$.
For general $m$ you can use the generalized Leibniz rule. This calculation is a bit more technical, but the idea of it is exactly the same as in the case $m=1$.
So it all comes down to $|e^{ix}| = 1$, since this makes the exponential terms vanish in our calculations.