Show that $V(Y-X^2)$ is irreducible. (Proof check) First, let me say that I know that there is another post on this very same question but I just want someone to check my proof, which is different from the answer presented in that post so this isn't a duplicate.
Now for my proof:
I proceed by showing that $I(V(Y-X^2))$ is prime. 
Assume $F \in I(V(Y-X^2))$ then $G(Z)\in \mathbb{C}[Z]$ defined by $G(Z) = F(Z,Z^2)$ then we have that $G(Z) = 0$ for all $Z \in \mathbb{C}$ thus $G(Z) = 0$. Now assume $F = fg$ where $f\notin I(V(Y-X^2))$, then define $f' = f(Z,Z^2)$ and $g' = g(Z,Z^2)$ then $f'g' = 0$, but we know that $f' \not= 0$ thus $g'=0$ and so $I(V(Y-X^2))$ is prime.
something seems off about this proof, and if true then this proof can easily be generalized to prove that a huge class of algebraic sets are irreducible, so I'm a bit suspicious. 
 A: I can't find a flaw in this proof (I believe it's correct), but I don't think it'll generalize to a large class of polynomials in the way you think. The proof you give is implicitly relying on the fact that $F\in I(V(Y-X^2))$ if and only if $F(z,z^2)=0$ for all $z\in\Bbb C$, which has to do with the primality of $Y-X^2$ to begin with. For instance, at first glance it seems like the same argument should hold for the polynomial $Y^2-X^2$, but the problem lies in the fact that it's not true that $F\in I(V(Y^2-X^2))$ if and only if $F(z^2,z^2)=0$ for all $z\in\Bbb C$, or iff $F(z,z)=0$ for all $z$. It holds if and only if $F(z,z)=F(z,-z)=0$ for all $z$. Furthermore, the same argument won't work with the polynomial $1-X$, even though this is prime, and $V(1-x)$ is irreducible. It seems that the only polynomials that this argument could work for would be ones of the form $Y-X^n$ for some $n$, or $X-Y^n$.
It would likely be much easier to show in general that if $P$ is a prime ideal, then $V(P)$ is irreducible. Then what you want follows immediately because $(Y-X^2)$ is a prime ideal of $\Bbb C[X,Y]$.
