Evaluate the index of a vector field at an isolated zero, using the Brouwer degree I want to find the index of v(x,y)=(x,-y) at (0,0), which is equal to the degree of the mapping $f:S^1\to S^1$ defined by
$$f(x,y)=\frac{v(x,y)}{||v(x,y)||}.$$
To find $\deg(f)$, one needs to pick a regular value $\mathbf y$ and compute
$$\sum_{\mathbf x\in f^{-1}\ (\{\mathbf y\})}\mathrm{sgn}(\det(\mathrm{d}f_{\mathbf x})).$$
The choice of a regular value is irrelevant, and I choose $\mathbf y=(0,1)$, which gives $\mathbf x=(0,-1)$.
Here comes the problem. I don't know how to compute $\mathrm{d}f_{\mathbf x}$. Even if I know it's a linear transformation from $T_{\mathbf x} S^1$ to $T_{\mathbf y} S^1$, I still can't make it. Does it help to study the curve $(\cos t,\sin t)$? This curve passes through $\mathbf x$. Any advice is welcomed. Thanks. 
 A: The map $f:S^1\to S^1$ you defined is given by $f(x,y)=(x,-y)$. In complex notations, $f$ is just $f:z\to \bar{z}$. First you can notice that $f\circ f =id_{S^1}$, hence $1=\text{deg}(id)=\text{deg}(f\circ f)=\text{deg}(f)^2$, so that $\text{deg}(f)\in\lbrace 1,-1\rbrace$. We can show that f has degree $-1$.
To do so, we have to understand first what is $T_x S^1$. If we see $S^1$ as the level set $(h(x)=1)$ where $h(x)=\Vert x \Vert^2$, we have $$T_xS^1 = \text{Vect}(\nabla_x h)^{\perp}=\text{Vect}(x)^{\perp}$$
because $\nabla_x h = 2x$. Now we have to compute the sign of the determinant of the linear map $$d_x f:\text{Vect}(x)^{\perp}\to\text{Vect}(f(x))^{\perp}$$
for some $x\in S^1$. The "tricky" thing is that this sign depends of the base you choose for each of the two vector spaces in question. If $x\not = f(x)$, you have to specify an orientation on the circle before you compute $d_xf$ and the sign of its determinant. But this is really confusing, that is why I suggest that you compute the sign of the determinant of $d_xf$ at a point $x$ fixed by $f$, such as $x_0=(1,0)$. In this case, you have a map $d_{x_0}f:T_{x_0}S^1\to T_{x_0}S^1$ and you don't have to explicit an orientation on $S^1$. You just need to give the same base to $T_{x_0}S^1$ and $T_{x_0}S^1$ (the other one). Here you can take $e_1=(0,1)$ as a base for $T_{x_0}S^1$.
The last thing left to do is compute $d_{x_0}f(e_1)$. Because $f$ extends to a linear map $\tilde{f}:\mathbb{R}^2\to\mathbb{R}^2$, defined by $\tilde{f}(x,y)=(x,-y)$, we have $d_{x_0}\tilde{f}=\tilde{f}$ (with the good identifications) and so $d_{x_0}f(e_1)=d_{x_0}\tilde{f}(e_1)=\tilde{f}(e_1)=-e_1$. So $d_{x_0}f$ is just $-Id_{T_{x_0}S^1}$ which has determinant -1. Finally $$\text{ind}_{(0,0)}(v)=-1.$$
I hope this is helpful! I don't know if is is possible to prove this using the curve you mentioned. I think that you can do a nice "proof" of this by drawing a vector field on a surface $M$ with $\chi(M)=-1$ having only one zero, such that the vector field looks like $v$ around that $0$, and the Poincaré-Hopf theorem would give you the answer.
