Is it needed to increase the precision of ContourPlot in this case?

I have this function and I want to see where it is zero. $ $ \frac{1}{16} \left(\sinh (\pi x) \left(64 \left(x^2-4\right) \cosh \left(\frac{2 \pi x}{3}\right) \cos (y)+\left(x^2+4\right)^2+256 x \sinh \left(\frac{2 \pi x}{3}\right) \sin (y)\right)+\left(x^2-12\right)^2 \sinh \left(\frac{7 \pi x}{3}\right)-2 \left(x^2+4\right)^2 \sinh \left(\frac{5 \pi x}{3}\right)\right)+2 \left(x^2-4\right) \sinh \left(\frac{\pi x}{3}\right)$ $ I use ContourPlot

f[x_, y_] :=    2 (-4 + x^2) Sinh[(π x)/3] +     1/16 (((4 + x^2)^2 + 64 (-4 + x^2) Cos[y] Cosh[(2 π x)/3] +           256 x Sin[y] Sinh[(2 π x)/3]) Sinh[π x] -        2 (4 + x^2)^2 Sinh[(5 π x)/3] + (-12 + x^2)^2 Sinh[(         7 π x)/3]);  ContourPlot[  f[x, y] == 0, {x, 3.465728, 3.465729}, {y, 1.046786, 1.046795},   PlotPoints -> 500]  

and I obtain this plot

enter image description here

Now, my question is that can I trust this plot and conclude that the curves do not cross?

Or, I should increase the precision of the plot? And if so, how can I ask Mathematica to give higher precision for the axis in ContourPlot?