For $1\leqslant p\leqslant \infty$, $\Omega$ an open set, define the homogeneous Sobolev space $$\dot{W}^{1,p}(\Omega)=\{u\in L_{loc}^1(\Omega): D^\alpha u\in L^p(\Omega),\ |\alpha|=1\}.$$
Let $\varphi\in C_c^\infty(\Omega)$ which is the smooth function with compact support. My question is, given $f\in \dot{W}^{1,\infty}(\Omega)$, whether we can get $f\varphi\in \dot{W}^{1,\infty}(\Omega)$ or not.
Yes, this is true. The proof is below.
Beware nevertheless that you cannot have an inequality of the form $$ | f \varphi |_{\dot{W}^{1,\infty}(\Omega)} \leq \| \varphi \|_{C^1} \cdot |f|_{\dot{W}^{1,\infty}(\Omega)} \quad \text{i.e.} \quad \| D^\alpha ( f \varphi ) \|_{L^\infty(\Omega)} \leq \| \varphi \|_{C^1} \| D^\alpha f \|_{L^\infty(\Omega)} $$ because you might have $f(x) = C$ a constant, for which the right-hand sides are $0$, but the left-hand sides might be arbitrarily large.
Now the proof. First, $f \varphi \in L^1_{loc}(\Omega)$ because $f \in L^1_{loc}(\Omega)$ and $\varphi \in L^\infty(\Omega)$. Moreover, from the Leibniz formula: $$ D^\alpha(f\varphi) = (D^\alpha f) \varphi + f (D^\alpha \varphi). $$ The first term is easy: $D^\alpha f \in L^\infty(\Omega)$ was the assumption, and $\varphi \in L^\infty(\Omega)$ also, so their product too. For the second term, since $D^\alpha \varphi \in L^\infty(\Omega)$ and is compactly supported, we only need to check that $f \in L^\infty_{loc}(\Omega)$.
Let $K \subset \Omega$ be a compact set. Up to a finite covering, we can assume that $K$ is a closed ball of radius $r > 0$. We want to prove that $f \in L^\infty(K)$. Taking any pair of points $a,b$ in the ball, we have $$ |f(b)-f(a)| \leq 2r \|D f\|_{L^\infty(K)} $$ hence $$ |f(b)| \leq |f(a)| + 2r \|D f\|_{L^\infty(K)}. $$ Averaging over $a \in K$, we obtain $$ |f(b)| \leq C \| f \|_{L^1(K)} + C\|D f\|_{L^\infty(K)}. $$ Since $f \in \dot{W}^{1,\infty}(\Omega)$, the right-hand side is finite, so $f \in L^\infty(K)$. Hence $f \in L^\infty_{loc}(\Omega)$, which concludes the proof.