I don't understand what I'm missing in this example.
Let $X=V(X_1^2+X_2^2-X_0^2)$ the circle in $\mathbf{P}^2_k$, being $k$ an algebraically closed field. Let be also $f:X\longrightarrow \mathbf{P}_k^1$ the regular map defined by $f(z_0 :z_1 :z_2)=[z_1 :z_2]$.
How can I compute $f^* (k(\mathbf{P}^1_k))$ in $k(X)\simeq k(t)$?
I noted that for every rational function $\varphi\in k(\mathbf{P}^1_k)$ we have $$ f^*\varphi (z_0:z_1:z_2)=\varphi (z_1:z_2)$$ ("switching the variable") but I can't see how this can help me.
Edit: I corrected the above mistake; $k(\mathbf{P}^1_k)$ is $k(t)$, the quotient field over $k[t]$, where $t$ is an indeterminate.
Since you didn't get an answer yet, let me try to give one. I hope it is helpful.
First of all, define the following rational functions on $X$:
$$ x_0 = \frac{X_0}{X_2}, \quad x_1 = \frac{X_1}{X_2}.$$
Then $k(X)= k(x_0,x_1)$, and these two generators satisfy the equation $x_1^2-x_0^2=1$.
For your description $k(X)=k(t)$, we need to choose an appropriate generating rational function $t$: one choice is $t=x_0+x_1$. Then using the above equation we find that $\frac{1}{t}=x_1-x_0$, and hence $x_1=\frac{1}{2}(t+\frac{1}{t})$.
Now if $Y,Z$ are homogeneous coordinates on $\mathbf P^1$, then $k(\mathbf P^1) = k(\frac{Y}{Z})$. So your formula for the map $f$ shows that $f^*(k(\mathbf P^1))=k(\frac{X_1}{X_2})=k(x_1)=k(t+\frac{1}{t})$.
Finally one should check that the extension $k(t+\frac{1}{t}) \subset k(t)$ really is degreee 2. For this, write $t+\frac{1}{t}=s$; then the minimal polynomial of $t$ with coefficients in $k(s)$ is $T^2-sT+1$.