= Solution
Suppose first that the <affine algebraic set>[affine variety] $V$ is irreducible. If $fg\in I(V)$, then
$$
V=(V\cap V(f))\cup(V\cap V(g)).
$$
Both pieces are <Zariski-closed set>[Zariski closed] in $V$. Irreducibility forces one to equal $V$, so either $f\in I(V)$ or $g\in I(V)$. Thus $I(V)$ is a <prime ideal>.
Conversely, suppose that $I(V)$ is prime and that $V=A\cup B$ with $A,B$ closed in $V$. If both were proper, the <Hilbert Nullstellensatz> would give
$$
I(V)\subsetneq I(A),
\qquad
I(V)\subsetneq I(B).
$$
Choose $f\in I(A)\setminus I(V)$ and $g\in I(B)\setminus I(V)$. Their product vanishes on $A\cup B=V$, so $fg\in I(V)$, contradicting primality. Hence one piece equals $V$. This proves the criterion of <irreducibility and prime vanishing ideal>:
$$
\boxed{V\text{ is irreducible}\iff I(V)\text{ is prime}}.
$$
For the <semicubical parabola>
$$
C=V(y^2-x^3),
$$
consider
$$
\theta:k[x,y]\to k[t],
\qquad
\theta(x)=t^2,\quad\theta(y)=t^3.
$$
Certainly $(y^2-x^3)\subseteq\ker\theta$. Divide any polynomial by the monic polynomial $y^2-x^3$ as a polynomial in $y$; its remainder is $a(x)y+b(x)$. If this remainder lies in $\ker\theta$, then
$$
a(t^2)t^3+b(t^2)=0.
$$
The first summand contains only odd powers of $t$ and the second only even powers, so both vanish and $a=b=0$. Therefore
$$
\ker\theta=(y^2-x^3),
\qquad
k[C]\cong k[t^2,t^3].
$$
The latter is a subring of the <integral domain> $k[t]$, hence is a domain. Thus $(y^2-x^3)$ is prime and
$$
\boxed{C\text{ is irreducible}}.
$$
An affine variety of dimension $d$ is smooth at $p$ when its <Zariski tangent space> has dimension $d$; equivalently, the <Jacobian criterion> requires the Jacobian of its defining ideal to have rank $n-d$. For this plane curve, put $F(x,y)=y^2-x^3$. Its gradient is
$$
\nabla F=(-3x^2,2y).
$$
At $(0,0)\in C$ the gradient vanishes, so the tangent space is all of $k^2$ and has dimension two rather than the curve dimension one. Hence the result on the <coordinate ring and singularity of the semicubical parabola> gives
$$
\boxed{C\text{ is not smooth; its origin is singular}}.
$$
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