I came across the following example when reading Silverman:
Example 3.7 Let $V$ be the variety
$$V : Y^2Z = X^3 + X^2Z$$
and consider the rational maps
$$\psi : \mathbb{P}_1 → V, \psi = [(S^2 − T^2)T, (S^2 − T^2)S, T^3],$$ $$\varphi : V → \mathbb{P}_1, \varphi= [Y,X].$$
Here $\psi$ is a morphism, while $\varphi$ is not regular at $[0, 0, 1]$. Not coincidentally, the point $[0, 0, 1]$ is a singular point of $V$; see (II.2.1). We emphasize that although the compositions $\varphi \circ \psi$ and $\psi \circ \varphi$ are the identity map wherever they are defined, the maps $\varphi$ and $\psi$ are not isomorphisms, because $\varphi$ is not a morphism.
It seems to me that although $\varphi$ is not a morphism, the two maps defined above imply that $\mathbb{P}_1$ is birational to $V$ (since they define isomorphisms between $\mathbb{P}_1$ and $V\[0,0,1]$). On the other hand, $K(\mathbb{P}_1)=K(X)$ is purely transcendental, but $K(V)\cong K(X, \sqrt{X^3+X^2})$ is not, so $\mathbb{P}_1$ cannot be birational to $V$.
I would appreciate it if someone could point out the mistake in my argument. Thanks!