I am given the following polynomial $$f(x,y)=y^2-4x^3-g_2 x, $$ for a fixed constant $g_2$. A base point of $f$ is a point $(x_0,y_0)$ which is a member of all level curves $\{ (x,y): f(x,y)=c \},c \in \mathbb R$. Obviously the polynomial has no base points when considered in $\mathbb R^2$. Thus I was told to consider it in the projective space $\mathbb{CP}^2$. I'm not entirely sure what it means. I guess I should embed the point $(x,y) \in \mathbb{R}^2$ as $[1:x:y] \in \mathbb{CP}^2$, but I'm not really sure how to take it from there.
Thanks.
Yes, you're on the right track.
The projective closure of the zero set of the polynomial $f(x,y)-c$ is the zero set of the "homogenized" version of $f(x,y)-c$, that is the homogeneous polynomial
$$F_c(X,Y,Z)=Y^2Z-4X^3+g_2XZ^2+cZ^3$$
where $X,Y,Z$ are homogeneous coordinates on $\mathbf P^2$. Now you are trying to find points $[a,b,c] \in \mathbf P^2$ which lie in the set $F_c=0$ for all $c$.
As you have already observed, there are no such points in the "usual" $(x,y)$-plane; in projective terms this translates to the statement that no point $[\alpha,\beta,1]$ lies in $F_c=0$ for all $c$. So you can restrict attention to points $[\alpha,\beta,0]$. (Remember that in projective space, we can rescale coordinates by nonzero scalars, so that every point is represented by a triple of homogeneous coordinates in which either $Z=1$ or $Z=0$.)
But now setting $Z=0$ in the formula for $F_c$, we see that if $[\alpha,\beta,0] \in \{ F_c=0\}$ then we must have $\alpha=0$ too; rescaling $\beta$ (which must be nonzero) we get just one point, namely $[0,1,0]$. Conversely, it's a straightforward check to see that $[0,1,0] \in \{F_c=0\}$ for all $c$.
So there's exactly one base point in $\mathbf P^2$, namely the point $[0,1,0]$. (From a fancier point of view, we might view this as 9 "coincident" basepoints, or one point "with multiplicity 9", but that's a story for another day.)