Show that the meromorphic differential of the homogeneous polynomial is holomorphic and not isomorphic to $\mathbb{P_1}$ Consider the elliptic curve i.e. non-singular cubic, $X$ given by the equation $\xi_0\xi_2^2=\xi_1^3-\xi_0^2\xi_1$ in projective coordinates $(\xi_0:\xi_1:\xi_2)$, or, equivalently, by the equation $y^2=x^3-x$ in the affine coordinates $(x=\frac{\xi_1}{\xi_0},y=\frac{\xi_2}{\xi_0})$. Show that the meromorphic differential $\omega:=\frac{dx}{y}$ is holomorphic, i.e., has no poles and has no zeros. Explain why $X$ is not isomorphic to $\mathbb{P_1}$. Show that the genus of $X$ is $g=1$.
Attempt:
First consider the point at infinity $(0:0:1)$. Then $v_1=0$ and $v_2=0$ for the affine coordinates $v_1=\frac{\xi_0}{\xi_2}$ and $v_2=\frac{\xi_1}{\xi_2}$. These are substituted into the homogeneous equation above to give $G(v_1,v_2)=-v_1+v_2^3-v_1^2v_2=0$. And so,
$\frac{\partial G}{\partial v_1}dv_1+\frac{\partial G}{\partial v_2}dv_2=(-2v_1v_2-1)dv_1+(3v_2^2-v_1^2)dv_2=0$.
$\implies dv_1=\frac{3v_2^2-v_1^2}{2v_1v_2+1}dv_2$
Now we have affine coordinates $x=\frac{\xi_1}{\xi_0}$ and $y=\frac{\xi_2}{\xi_0}$ for $\xi_0\not=0$. So $x=\frac{v_2}{v_1}$ and $y=\frac{1}{v_1}$. 
This gives us $\omega=\frac{dx}{y}=v_1d(\frac{v_2}{v_1})=v_1(\frac{dv_2}{v_1}-\frac{v_2}{v_1^2}dv_1)=dv_2-\frac{v_2}{v_1}dv_1= dv_2-(\frac{v_2}{v_1})(\frac{3v_2^2-v_1^2}{2v_1v_2+1})dv_2=(1+\frac{3v_2^3-v_1^2v_2}{2v_1^2v_2+v_1})dv_2=(\frac{3v_1^2v_2-3v_2^3+v_1}{2v_1^2v_2+v_1})dv_2$
But from above we know that $v_1=v_2^3-v_1^2v_2$ so,
$\omega=(\frac{2v_1^2v_2-2v_2^3}{v_1^2v_2+v_1})dv_2=(\frac{-2v_1}{2v_2^3-v_1})dv_2$.
When I get to this step I get stuck. I have been taught that I should try to get the function in terms of $v_1$ only, i.e. $\omega=g(v_1)dv_2$, and then set this equal to zero to show that there are no poles or zeros and hence the meromorphic differential is holomorphic. But I can't seem to reduce the last relation above to be in terms of only $v_1$.
As for why $X$ is not isomorphic to $\mathbb{P_1}$, I think it might have to do with the fact that there are no points in $X$ that correspond to infinity in $\mathbb{P_1}$? But I'm not sure. Any help with this problem is greatly appreciated. 
Thanks.
 A: The form $\omega$ is clearly holomorphic except maybe at the three points at a finite distance $P_i=(x_i,0)$ where $y=0, x_i=0,1,-1$ or at the point $Q_\infty=(0:0:1)$ .     
1) At a finite distance we write  $y^2=x^3-x$ so that $2ydy=(3x^2-1)dx$ and thus $\omega= \frac {dx}{y}=\frac{2dy}{3x^2-1}$ which has  neither zero nor pole at $P_i$.  
2) At infinity we choose coordinates $u=\frac {\xi_0}{\xi_2}, v=\frac {\xi_1}{\xi_2}$ and the curve has equation $u=v^3-u^2v$.   
We may take $v$ as uniformizing parameter at $Q_\infty$ and then $du=\frac{3v^2-u^2}{1+2uv}dv$, as you correctly calculated yourself. Then $$\omega=\frac {dx}{y}=ud(\frac vu)=dv-\frac vu du=dv-\frac vu \frac{3v^2-u^2}{1+2uv}dv=\frac{u+3u^2v-3v^3}{u(1+2uv)}dv=\frac{u+3u^2v-3(u+u^2v)}{u(1+2uv)}dv$$ so that $\omega =\frac{-2}{1+2uv}dv$ is holomorphic without zero also at the point $Q_\infty$ with  coordinates $u=v=0$.  
Conclusion
Since $X$ has a non-zero global holomorphic differential form $\omega$ and $\mathbb P^1$ has none, the curves $X$ and  $\mathbb P^1$ are not isomorphic.  [Actually they are not even homeomorphic].
