Computation of contour integral without winding numbers.

I am trying to solve problem $ 3$ on page 108 in Ahlfors. The problem asks to compute $ \int_{\left\vert z \right\vert = 2} \frac{dz}{z^2 – 1}$ so I am trying to do so without the use of winding numbers, which isn’t introduced until later sections.

I first used partial fraction decomposition to write $ \frac{1}{z^2 – 1}$ as $ \frac{1}{2} \cdot (\frac{1}{z-1} – \frac{1}{z+1})$ . From here, it’s clear that the integral is $ 0$ if we resort to winding numbers. Since I was unable to compute the integral using elementary methods, I resorted to reading this solution. However, I don’t understand how they changed the path of integration from $ \left\vert z \right\vert = 2$ to $ \left\vert z – 1 \right\vert = 1$ because when $ z = -2$ , $ \left\vert z – 1 \right\vert = \left\vert -3 \right\vert = 3 \ne 1$ .

Calculating curvature of a contour

I have a equation of a scalar field in the form


I want to find the curvature of the contour of the curve at fc=f(0.5,0.5).

So I need to calculate the derivative dy/dx and d/dx(dy/dx)

I can solve the equation f(x,y)=fc and get the derivative of f(x,y) w.r.t x

On paper we do,

d/dx (f(x,y))=d/dx(fc) 2*x+2*y*(dy/dx)+y+x(dy/dx)=0 (dy/dx)=-(2*x+y)/(x+2*y) 

and further d/dx(dy/dx) for a curvature approximate

how can I rearrange the equation such that I can get the value of dy/dx on mathematica