I have a question about the computation of global regular function of $\mathbb{P}^2-V(x_0^2+x_1^2+x_2^2)$. I konw this is isomorphic to an affine variety, and i tried to compute the ring of global regular functions by Veronese map, but i'm not sure it is right. By Veronoese map $\mathbb{P}^2$ is mapped in $\mathbb{P}^5$ by $$[x_0,x_1,x_2] \rightarrow [x_0^2,x_1^2,x_2^2,x_0x_1,x_0x_2,x_1x_2].$$ In particular the conic is (say $[t_0,...,t_5]$ coordinates of $\mathbb{P}^5$) given by $t_0+t_1+t_2=0$ (intersect with the image of $\mathbb{P}^2$).
Now $\mathbb{P}^2-V(x_0^2+x_1^2+x_2^2)=X\cap \{t_0+t_1+t_2\not=0\}$ where X is the image of $\mathbb{P}^2$ by Veronese map. Moreover changing the variables of P^5 i can suppose that $$[x_0,x_1,x_2] \rightarrow [x_0^2,x_1^2,x_2^2+x_1^2+x_0^2,x_0x_1,x_0x_2,x_1x_2]$$ so my initial variety is given by points of the form $$(\frac{x_0^2}{x_0^2+x_1^2+x_2^2},\frac{x_1^2}{x_0^2+x_1^2+x_2^2},\frac{x_0x_1}{x_0^2+x_1^2+x_2^2},\frac{x_0x_2}{x_0^2+x_1^2+x_2^2},\frac{x_1x_2}{x_0^2+x_1^2+x_2^2})$$ in $\mathbb{A}^5$. I would to understand which variety is isomophic to it (it seems to be an hypersurface in $\mathbb{A}^3$) in such way to determine easily the ring of global regular functions. Maybe there is a faster way? or a different reasoning to conclude? Thanks for the help.