I have to show that the variety $$C:=\{(x_0 :x_1:x_2)\in \mathbb{P}_k^2\ |\ x_0^2x_1^2 + x_0^2x_2^2+x_1^2x_2^2 = 0\}\subset \mathbb{P}_k^2$$ is rational.
I already found a solution on this but I don't understand it completely. The solution says that we can use the Cremona-transformation $$\phi: \mathbb{P}_k^2 \to \mathbb{P}_k^2,\ (x_0:x_1:x_2)\mapsto (\frac{1}{x_0}:\frac{1}{x_1}:\frac{1}{x_2})=(x_1x_2:x_0x_2:x_1x_2),$$ which is birational with $\phi=\phi^{-1}.$ Therefore, $\phi^*: k(\mathbb{P_k^2)}\to k(\mathbb{P_k^2})$ is an isomorphism between the function-fields.
It is $\phi^*(x_0^2x_1^2 + x_0^2x_2^2+x_1^2x_2^2) = x_2^2 + x_0^2 + x_1^2$, now the solution says it holds that $C\cap\{x_0x_1x_2 \neq0\}$ is isomorph to $Q\cap\{x_0x_1x_2 \neq0\}$, where $Q=\{x_0^2+x_1^2+x_2^2=0\}$ is a smooth conic section, which is known to be isomorph to $\mathbb{P_k^1}$.
I understand that $\phi^*$ is an isomorphism and that $\phi^*(x_0^2x_1^2 + x_0^2x_2^2+x_1^2x_2^2) = x_2^2 + x_0^2 + x_1^2$. But why can I follow that the two varietys, induced by those polynomials, are isomorph?