Let $\varphi \in \mathcal{D}(\mathbb{R}^n)$ and $h \in \mathbb{R}^n \setminus \{0\}$. For $t \in \mathbb{R}^n \setminus \{0\}$ we put $$ \varphi_t(x)= \dfrac{\varphi(x+th)-\varphi(x)}{t}. $$ The question is: prouve that $\varphi_t \in \mathcal{D}(\mathbb{R}^n)$ for all $t \neq 0$.
My purpose is: Let $t \neq 0$. We have $\varphi_t \in C^{\infty}(\mathbb{R}^n)$ beacause $\varphi \in \mathcal{D}(\mathbb{R}^n)$.
Now, it remains to prove that $Supp \varphi_t$ is compact. We have $Supp \varphi_t \subset Supp(x \to \varphi(x+th)) \cup Supp \varphi$, but the problem is that $Supp(x \to \varphi(x+th)) \cup Supp \varphi$ depends on $t$ and it seems to me that the $Supp \varphi_t$ must to be independant on $t$.
My questions are: please, how we find $Supp \varphi_t$ independent on $t$? And why the support have to be independent on $t$?
If I understood correctly the question, you are asked to prove that for any fixed $t\in\mathbb R^n\setminus\left\{ 0\right\} $, the function $\varphi_t$ belongs to $\mathcal D\left(\mathbb R^n\right)$. Therefore, fix once for all $t$.
You can use the fact that for any two functions $f$ and $g$, $$\operatorname{Supp}\left(f+g\right) \subset \operatorname{Supp}\left(f\right)\cup \operatorname{Supp}\left(g\right) $$ (because if $x$ does not belong to the right hand side, we can find a small ball around $x$ such that $f$ and $g$ vanish on that ball, hence so does $f+g$).
The support of the function $x\mapsto \varphi\left(x+th\right)$ is compact (recall that $t$ and $h$ are fixed hence the support is only a translate of the support of $\varphi$) hence the support of $\varphi_t$ is bounded. Since it is by definition closed, this ends the proof that $\varphi_t$ belongs to $\mathcal D\left(\mathbb R^n\right)$.