Hi Diane!

This is not much but I hope this can help. This is for the first part without using involutions. What I've done so far is the following:

First represent the curve

where and span the plane containing the curve and .

It's like a change of basis from { , } to {(1, 0, 0), (0, 1, 0)}

Then we can take the curve

where , , , and are nonzero constants (this is to ensure that we get a distinct curve). This curve has the same domain as and observe that if we differentiate this, we get

Hence, as desired.

If you consider the function

,

It can be verified that this is a bijection since is well-defined and since has the same domain as .