Let us remember how the space of Schwartz. A multi-index $\alpha=(\alpha_1,...,\alpha_d)$ is an element of $\mathbb{N}_{0}^d$, where $\mathbb{N}_0=\mathbb{N}\cup \{0\}$, and the order of the multi-index $\alpha$ is the number $$|\alpha|=\sum_{j=1}^d\alpha_j.$$ Additionally we introduce the following notations: $$x^\alpha:=x_{1}^{\alpha_1}x_{2}^{\alpha_2}\cdots x_{d}^{\alpha_d},$$ $$\partial^\alpha:=\partial_{x_1}^{\alpha_1}\partial_{x_2}^{\alpha_2}\cdots \partial_{x_d}^{\alpha_d},\quad \mbox{where}\hspace{.1cm} \partial_{x_i}^{\alpha_i}\hspace{.1cm} \mbox{denotes the operator}\hspace{.1cm}\frac{\partial^{\alpha_i}}{\partial x_{i}^{\alpha_i}},$$ $$f^{(\alpha)}:=\partial^\alpha f,$$ provided that the derivatives of $f$ exist. In case that $x_i=\alpha_i=0$ for some $i\in\{1,...,d\}$, for convenience we will assume that in the expression $x^\alpha$ the term $x_{i}^{\alpha_i}$ is equal to 1.
Definition (Schwartz space). The Schwartz space, denoted by $\mathcal{S}(\mathbb{R}^d)$, is the set formed by all functions $f\in C^\infty(\mathbb{R}^d)$ such that $$\|f\|_{\alpha,\beta}:=\displaystyle{\sup_{x\in\mathbb{R}^d}}\left|x^\alpha f^{(\beta)}(x)\right|<\infty\qquad\mbox{for all}\hspace{.1cm} \alpha, \beta\in\mathbb{N}_{0}^d.$$ The Schwartz space is also known as the space of infinitely differentiable functions of rapid decrease.
Below I present the demonstration that I have done of the exercise in question seeking the approval of the community, since I have some doubts about the proof.
With the definition clear, we proceed to demonstrate the exercise. Since $f\in C^\infty(\mathbb{R}^n)$, it is clear that each $\frac{\partial f}{\partial x_j}\in C^\infty(\mathbb{R}^n)$ and $x_j f\in C^\infty(\mathbb{R}^n)$. We also have to $\left|x^\alpha \left(\frac{\partial f}{\partial x_j}\right)^{(\beta)}\right|$ is bounded, since otherwise $\|f\|_{\alpha,\beta}:=\displaystyle{\sup_{x\in\mathbb{R}^n}}\left|x^\alpha f^{(\beta)}(x)\right|$ it wouldn't be finite, contradicting that $f\in\mathcal{S}$ (It seems to me that it is easy to see this, but this is one of the reasons why I have posted this proof, since I do not know how to explain it more clearly). Therefore $$\displaystyle{\sup_{x\in\mathbb{R}^n}}\left|\frac{\partial f}{\partial x_j}\right|<\infty,$$ and hence $\frac{\partial f}{\partial x_j}\in\mathcal{S}$.
On the other hand, on the other hand, since $f\in \mathcal{S},$ $$\displaystyle{\sup_{x\in\mathbb{R}^n}}\left|x^\alpha f^{(\beta)}(x)\right|<\infty,$$ it follows that $\displaystyle{\sup_{x\in\mathbb{R}^n}}\left|x^\alpha x_jf^{(\beta)}(x)\right|<M|x_j|,\quad\mbox{para alguna constante}\hspace{.1cm} M,$ this is, $$\displaystyle{\sup_{x\in\mathbb{R}^n}}\left|x^\alpha x_jf^{(\beta)}(x)\right|<\infty.$$ It is concluded that $x_jf\in\mathcal{S}$.