I'm pretty sure it is a smooth point since given $f(x,y)=x-y^2$ the gradient $df=(1,-2y)$ is always non-singular. I'm asking because page 22 of Principle of Algebraic Geometry says:
Note that p is a smooth point of V if and only if mult_p(V)=1.
The multiplicity is defined to be the sheets of the covering map induced by the variety around p. Since 0 is a multiple root of $y^2$, (0,0) is a branch point of Riemann surface defined by $f$, so any neighborhood of (0,0) will have 2 sheets, which means (0,0) is singular based on the above statement.
What's wrong here?
As commented on by KReiser, $x-y^2$ does not have multiplicity $2$ at the origin. There is a complicated algebraic way of defining multiplicity (see the entire book Algebre Locale Multiplicities by Serre).
Intuitively, one can get at the multiplicity geometrically though. In our tame case, you can get the multiplicity of $x-y^2$ at $(0,0)$ by testing its intersections against lines. In particular, consider the line $\ell=\{y=0\}$. Then, the intersection of $\ell$ and $V(x-y^2)$ has multiplicity $1$ at $(0,0)$.
This multiplicity is calculated as
$$\ell_{k[x,y]_{(x,y)}} \,\,k[x,y]_{(x,y)}/(x-y^2,y)=1$$