Ever wondered what dtau/dx' means?
Turn it upside down....
dx'/dtau.
That looks a lot like dx/dt = v, doesn't it?
dt/dx =1/v

Now, since there can only be one v between two frames,
dtau/dx' = dt/dx.
1/v + [v/(c^2 - v^2)]dtau/dt = 0
v/(c^2 - v^2).dtau/dt =  -1/v      <---- note carefully, because v = 0 as  I
have repeatedly told you.

(c^2 - v^2).dtau/dt =  -1
dtau/dt =  -1/(c^2 - v^2)
           =  -1/(1 - v^2/c^2)

What happened to the square root, nTaul, and why the minus sign?
It's amazing what results you can get when you divide by zero, isn't it?
Did you know that division by zero is undefined, nTaul?