The image given is only an embedding of manifolds, and doesn't look like an embedding of differential manifolds, and even less of an embedding of smooth manifolds.
To be an embedding of differential manifolds, you need your map to induce an injective map on tangent vectors. Here, at the end of the infinitely shrinking knot, there is no tangent line to your embedding : on one side you want to send the tangent vector to a horizontal vector, but on the other side, the slope keeps turning in circle as you get closer to the singularity.
However, all is not lost, you can squeeze the chain : instead of using a map of the kind $t \mapsto (tx(\log t), ty(\log t), tz(\log t))$ for some periodic functions $x,y,z$, use $t \mapsto (tx(\log t), t^2x(\log t)y(\log t), t^2x(\log t)z(\log t))$ instead. It is still not differentiable at $t=0$ because $x(\log t)$ isn't convergent, but at least there is a tangent line. I think you may find a differentiable wild knot with enough tinkering.
But you can't have a map of $C^1$ manifolds (nor of smooth manifolds) that looks like that. If the map was $C^1$, the tangent line has to vary continuously when you move along the knot, so there has to be a whole neighboorhood around the singularity where all the tangent lines don't differ from the horizontal line by, say, a $\pi/4$ angle. But you can't make any wild knotting with only using tangent lines in that cone. In particular, you can locally thicken your knot by attaching small disks all orthogonal to the original horizontal tangent vector.
Then since the circle is compact, you should be able to get a global thickening from a finite number of those local ones.