Find and show that the residues of the meromorphic differential $dx$ for Fermat's curve $C:$ $\xi_1^4+\xi_2^4=\xi_0^4$ is zero Find the residues of the meromorphic differential $dx$ of Fermat's curve $C:$ $\xi_1^4+\xi_2^4=\xi_0^4$ at its poles. Check that their sum is zero, as it must be.
Attempt:
Let $\xi_2\not=0$. Then the corresponding affine coordinates are $v_0=\frac{\xi_0}{\xi_2}$ and $v_1=\frac{\xi_1}{\xi_2}$. So in affine coordinates the curve is $G(v_0,v_1)=v_1^4-v_0^4+1=0$.
So $\frac{\partial G}{\partial v_0}dv_0+\frac{\partial G}{\partial v_1}dv_1=(-4v_0^3)dv_0+(4v_1^3)dv_1=0 \implies dv_1=\frac{v_0^3}{v_1^3}dv_0 $.
Now, we have a transformation between affine coordinates. $x=\frac{\xi_1}{\xi_0}=\frac{v_1}{v_0}$. ($x$ is one of the affine coordinate for when $\xi_0\not=0$)
So $\omega=dx=d(\frac{v_1}{v_0})=\frac{1}{v_0}dv_1-\frac{v_1}{v_0^2}dv_0=\frac{1}{v_0}(\frac{v_0^3}{v_1^3})dv_0-\frac{v_1}{v_0^2}dv_0=\frac{v_0^4-v_1^4}{v_1^3v_0^2}dv_0$
But we know $v_0^4-v_1^4=1$ so,
$\omega=\frac{1}{v_0^2v_1^3}dv_0$
At this point I do not know what to do next. I believe my function is supposed to have degree of $-4$ but the equation above is degree $-5$. Is there a zero somewhere that is "hiding" from me in the form I currently have?
Any help would be greatly appreciated. Thanks.
 A: The four points at infinity of $C$ are $(\xi_0:\xi_1:\xi_2)=P_i=(0:a_i:1)$ where the $a_i$'s are the four complex roots of $z^4+1=0$.
In the affine coordinates $(v_0,v_1)$ we have $$v_1^4-v_0^4+1=0,\quad v_1^3dv_1-v_0^3dv_0=0\quad (\bigstar)$$ and our points at infinity have coordinates $P_i=(v_0,v_1)=(0,a_i)$.    
At $P_i$ the implicit function theorem allows us to take $v_0$ as uniformizing parameter (=coordinate) for the curve $C$ because  $\frac {\partial }{\partial v_1}(v_1^4-v_0^4+1)\mid_{P_i}=4v_1^3\mid_{P_i}=4a_i^3\neq 0$ .
Since $\omega=
\frac{v_0^2}{v_1^3}dv_0-\frac{v_1}{v_0^2}dv_0$ as you quite correctly computed, the residue of $\omega$  at $P_i$ is $$\operatorname {Res}_{P_i}(\omega)=\operatorname {Res}_{P_i}(-\frac{v_1}{v_0^2}dv_0)$$ Indeed $\operatorname {Res}_{P_i}(\frac{v_0^2}{v_1^3}dv_0)=0$ because  $\frac{v_0^2}{v_1^3}dv_0$ is holomorphic at $P_i$ since $v_1$ doesn't vanish at $P_i$.
On the other hand $\operatorname {Res}_{P_i}(-\frac{v_1}{v_0^2}dv_0)=-\frac{dv_1}{dv_0}\mid _{P_i}=0$ too because $\frac{dv_1}{dv_0}=\frac{v_0^3}{v_1^3}$ as follows from $(\bigstar)$ .
This means that $\operatorname {Res}_{P_i}(\omega)=0$ and the total residue of $\omega$ is thus $$\operatorname {Res}(\omega)=\sum_{i=1}^4\operatorname {Res}_{P_i}(\omega)=0+0+0+0=0$$ just as it should!
