In the book Algebraic Curves and Riemann Surfaces by Rick Miranda, the author often makes use of the projection map $\pi \colon X \to \mathbb{P}^1$ given by $[x \colon y \colon z] \mapsto [x \colon z]$, where $X$ is a smooth projective plane curve. Why is such a map always well defined?
An example that shows it is not always possible to project to exactly this pair of coordinates is the curve given by $F(x, y, z) = x^2 -yz$, which contains the point $[0 \colon 1 \colon 0]$. But in this case it is possible to project onto $[y \colon z]$. Why is there always such a pair of coordinates?
You can certainly have curves that contain $[1 : 0 :0]$, $[0:1:0]$, $[0:0:1]$.
(Given a curve $C$ in the plane, with 3 points on the curve in general position - i.e. whose vectors are linearly independent -- you can find an automorphism of projective space that moves those points to $[1 : 0 :0]$, $[0:1:0]$, $[0:0:1]$. This is possible as long as $C$ isn't a line.)
What is true is that you can always find a point $P$ not on your curve, and a line $L$ that doesn't contain $P$, and project from $P$ to $L$. Equivalently, you can do a change of coordinates that allow you to project via $[x:y:z] \to [x:z]$.
Harris's book on algebraic geometry has nice pictures of this. You can also try looking up stuff related to 'projection from a point'.