This is part of exercise 16.5.A in Vakil's notes (http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf). To give an example of why the assumptions in the curve-to-projective extension theorem (16.5.1) are necessary, he asks you to show the following morphisms $C \backslash p \to Y$ do not extend to morphisms $C \to Y$:
$C = Y = \mathbb{A}^1_k$, $p=0$, and the map is $t \mapsto t^{-1}$.
$C = \mathbb{A}^2_k$, $Y = \mathbb{P}^1_k$, $p=(0,0)$, and the map is $(x,y) \mapsto [x : y]$.
- Same as 2 but $C = \mathrm{Spec}(k[x,y]/(y^2-x^3))$.
I completely understand the first one, and intuitively why the second and third should fail, but don't know how to prove it. Any hints are appreciated.
As 2) has been adressed in the comments, this answer will address 3).
Consider the function on the image of $C$ given by $[x:y]\mapsto y/x$. If the morphism extended to all of $C$, $y/x$ would be a function in $k[x,y]/(y^2-x^3)$, which is clearly wrong.
What's going on here is that singular curves are non-normal, and we're picking an element which is integral over the coordinate algebra but not contained in it - this is only possible when the curve is not normal.