Here is a picture of F (the arrows are scaled, to make the picture clearer).
Now look at the level curves of the two components of F1. These components are P1=, Q1=-2xy.
Their level curves are supposed to be perpendicular to each other. We first show P1, then Q1, then both together.
You can see the curves are perpendicular.
Example 2: Start with f(x)=Cos[x]. It turns out that Cos[x-iy]=Cos[x]Cosh[y]+iSin[x]Sinh[y],
so P3=Cos[x]Cosh[y], Q3=Sin[x]Sinh[y], and F3=(Cos[x]Cosh[y],Sin[x]Sinh[y]).
Here is a picture of F3:
Now F3=∇(Sin[x]Cosh[y]), and the functions
satisfy the Cauchy-Riemann equations, so, as with F1, the level curves of Q4 are the flow lines of F3.