Yes I agree withLet summarize the above comment of Hannes,comments there: In order for the seminorm here to be finite, one needs at least $|f(x)-f(y)| = o(|x-y|)$ when $x-y\to 0$, and this is thepossible only for constants functions. Since $B^1_{3,3}$ seminorn$f∈L^3$, $f=0$ (and so $f∈W^{1,3}$ ...).
I would addIf to avoid that one uses the second order difference $f(2y-x)-2f(y)+f(x)$ instead of $f(y)-f(x)$, one could however not conclude that $f∈W^{1,3}$, and this is due to the misleading definition of fractional Sobolev(-Slobodeckij) spaces $W^{s,p}$ since when $s>0$ is not an integer $W^{s,p} = B^s_{p,p} = F^s_{p,p}$ (where the $F^s_{p,q}$ are the Triebel-Lizorkin spaces) while $W^{n,p} = F^n_{p,2}$ when $n$ is an integer. An other fractional extension of Sobolev spaces are the Bessel-Sobolev spaces $H^{s,p}$ where the seminorm is the $L^p$ norm of the fractional Laplacian. They verify $H^{s,p} = F^s_{p,2}$ (For every $s≥0$).
All these spaces are ordered in this way when $p≥ 2$ (with strict inclusion when $p>2$) $$ B^s_{p,1} ⊂ B^s_{p,2} ⊂ F^s_{p,2} (=H^{s,p}) ⊂ F^s_{p,p} = B^s_{p,p} ⊂ B^s_{p,\infty}. $$
This is well explained for example in the book of Hans Triebel, Theory of Function Spaces II. Springer Basel, 1992.