$P^1$ not a regular level surface of a $C^1$ function on $P^2$

An other idea is to think of $P^1$ to be the subset in $P^2$ with homogeneous coordinates $[0:b:c]\in P^2$. In this way, you can see that $N=P^2\backslash P^1$ is exactly $\{[1:\alpha:\beta] \in P^2, \alpha,\beta\in \mathbb R \}$ which is clearly homeomorphic to $\mathbb R^2$ and hence connected.

Now suppose that $P^1$ is the regular level set of a smooth map $f:P^2\rightarrow \mathbb R$ , $P^1=f^{-1}(0)$.

By definition $f(N)$ lies in $\mathbb R\backslash \{0\}$, and since $P^2$ is a compact connected surface, one can assume that $f(P^2)=[0,1]$. Then every point in $f^{-1}(0)$ is a minimum for $f$ and has null differential which is a contradiction.