= Solution
Here <Morrey's inequality> has <Hölder exponent> $1-3/6=1/2$. Every $u\in W^{1,6}(\mathbb R^3)$ has a unique continuous representative $u^*$ such that
$$
\boxed{\|u^*\|_{C^{0,1/2}(\mathbb R^3)}\leq C\|u\|_{W^{1,6}(\mathbb R^3)},\qquad
[u^*]_{C^{0,1/2}(\mathbb R^3)}\leq C\|\nabla u\|_6.}
$$
We use the <Hölder norm> $\|f\|_{C^{0,1/2}}=\|f\|_\infty+[f]_{C^{0,1/2}}$, where the <Hölder seminorm> is $\sup_{x\ne y}|f(x)-f(y)|/|x-y|^{1/2}$. The second bound is the homogeneous form of <Morrey's inequality>; the first includes control of the values themselves.
Let $u_j=\rho_{1/j}*u$ be the <mollifications> from (i). They converge to $u$ in the <Sobolev norm>. Apply the assumed smooth version of <Morrey's inequality> to the differences $u_j-u_\ell$:
$$
\|u_j-u_\ell\|_{C^{0,1/2}}\leq C\|u_j-u_\ell\|_{W^{1,6}}\longrightarrow0.
$$
Thus $(u_j)$ is a <Cauchy sequence> in the complete <Hölder space> $C^{0,1/2}(\mathbb R^3)$ and converges there to a function $w$. In particular the convergence is uniform. On every <bounded set> $K$ of finite volume,
$$
\|w-u\|_{L^6(K)}
\leq |K|^{1/6}\|w-u_j\|_\infty+\|u_j-u\|_{L^6(\mathbb R^3)}\longrightarrow0.
$$
Exhausting $\mathbb R^3$ by <balls> proves $w=u$ <almost everywhere>. Set $u^*=w$. Taking limits in the smooth <Hölder norm> bound gives the first boxed estimate. For any fixed $x\ne y$, taking limits in
$$
|u_j(x)-u_j(y)|\leq C|x-y|^{1/2}\|\nabla u_j\|_6
$$
gives the second boxed estimate after taking the supremum. Finally, two continuous representatives equal <almost everywhere> must coincide everywhere, since a nonzero continuous difference would remain nonzero on an <open ball> of positive volume.
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