The curve has infinite length between $x=-1/\pi$ and $x=1/\pi$. To see this note that it passes through each point $$\left(\frac{2}{(4k+1)\pi},\frac{2}{(4k+1)\pi}\right)$$ just before passing through $$\left(\frac{2}{(4k+3)\pi},\frac{-2}{(4k+3)\pi}\right).$$ The distance between these two points is at least the absolute value of $\Delta y$, which is $$(2/\pi)\left[\frac{1}{4k+1}+\frac{1}{4k+3}\right]$$ which as $k \to \infty$ is asymptotic to $\frac{1}{k\pi}.$ [the same asymptotic estimate occurs between the points going with $1/(4k+3)\pi$ and $1/(4k+5)\pi.$]
So by limit comparison with the sum $\sum 1/(k\pi)$, and the fact that the length computed along straight line segments is less than the curve length, we see there is indeed infinite length as claimed.
This means when drawn the pen point would have to move with unbounded speed, or else that it would take an infinitely long time to draw.