Can a bidegree $(3,4)$ curve be embedded in plane? Suppose $C$ is a curve on $\mathbf{P}^1\times\mathbf{P}^1$ of bidegree $(3,4)$, why such a curve cannot be embedded as a curve in $\mathbf{P}^2$?

Projecting $C$ onto one of the factors gives you a $3:1$ ramified covering to $\mathbb P^1$, in other words a $g^1_3$ on $C$.
On the other hand the genus of $C$ is $(3-1).(4-1)=6$.
Now,  a plane smooth  curve in $\mathbb P^2$ can only have genus $6$ if it has a degree $d$ satisfying $\frac {(d-1).(d-2)}{2}=6$, in other words if it has degree $d=5$.
But a smooth plane curve of degree $5$ is not trigonal i.e. it has no $g^1_3$: see for example these notes by Harris, Page 45, Proposition 7.12.
(He actually proves that a smooth curve of degree $d$ cannot be a ramified covering of $\mathbb P^1$ with less than $d-1$ sheets)
Hence the trigonal curve $C$ cannot be embedded into $\mathbb P^2$.    
