= Solution
Choose a nonnegative <mollifier> $\rho\in C_c^\infty(\mathbb R^3)$ with $\int\rho=1$, and put $\rho_\varepsilon(x)=\varepsilon^{-3}\rho(x/\varepsilon)$. <Sobolev approximation by mollifiers> in the <Sobolev space> under consideration states
$$
\boxed{u_\varepsilon:=\rho_\varepsilon*u\in C^\infty(\mathbb R^3)\cap W^{1,6}(\mathbb R^3),\qquad
\|u_\varepsilon-u\|_{W^{1,6}}\longrightarrow0\quad(\varepsilon\downarrow0).}
$$
In particular, it gives <density of smooth functions in a Sobolev space>. We prove convergence of both the function and all its first <weak derivatives>.
The usual smoothing properties of <convolution> give $u_\varepsilon\in C^\infty$. For a <test function> $\phi\in C_c^\infty(\mathbb R^3)$ and $i\in\{1,2,3\}$, <Fubini's theorem> and the definition of a <weak derivative> give
$$
\begin{aligned}
\int u_\varepsilon(x)\partial_i\phi(x)\,dx
&=\int\rho_\varepsilon(z)\int u(y)\partial_i\phi(y+z)\,dy\,dz\\
&=-\int\rho_\varepsilon(z)\int\partial_i u(y)\phi(y+z)\,dy\,dz\\
&=-\int(\rho_\varepsilon*\partial_i u)(x)\phi(x)\,dx.
\end{aligned}
$$
The <integrals> are absolutely convergent: the <compact supports> of $\rho_\varepsilon$ and $\phi$ restrict $y$ to a <compact set>, on which $u$ and $\partial_i u$ are integrable by <Holder inequality>. Consequently
$$
\partial_i u_\varepsilon=\rho_\varepsilon*\partial_i u.
$$
The <approximate identity> theorem in the <Lp space> $L^6$ applies separately to $u$ and $\partial_i u$. Thus
$$
\|u_\varepsilon-u\|_6\to0,\qquad
\|\partial_i u_\varepsilon-\partial_i u\|_6\to0\quad(i=1,2,3).
$$
Each <convolution> belongs to $L^6$ by <Young's convolution inequality>. With the componentwise <Sobolev norm>, the sixth power of the required difference is the sum of these four sixth powers, proving the boxed claim. Any equivalent <Sobolev norm> gives the same convergence.
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