I'll use the more comfortable notations $w,x,y,z$ in place of $x_0,x_1,x_2,x_3$ and assume the base field has characteristic zero..
i) The curve $V$ is isomorphic to the plane curve $V'$ in the $x,y$ plane defined by $y^2=x^3$.
The isomorphism is $p:V\to V':(x,y,z)\mapsto (x,y)$ with inverse $p^{-1}:V'\to V:(x,y)\mapsto (x,y,x^3)$.
The curve $V'$ is irreducible because the polynomial $y^2-x^3$ is irreducible (or because it is the image of $\mathbb A^1\to \mathbb A^2: t\mapsto (t^2,t^3)$). Hence the isomorphic curve $V$ is irreducible too.
The only singularity of $V'$ is $(0,0)$ where the tangent space has dimension $2$ (immediate from the Jacobian criterion you mention) so that $V$ has $(0,0,0)$ as only singularity, with tangent space of dimension $2$.
ii) The surface $Y$ is irreducible because the polynomial $xy^2-z^3\in k[w,x,y,z]$ that defines it is irreducible (notice that it is of degree $1$ in $x$).
The absence of the variable $w$ in the equation indicates that $Y$ is the cone with vertex $[1:0:0:0]$ over the curve $C\subset \mathbb P^2_{x:y:z}=V(w)$ with equation $xy^2-z^3=0$.
The curve $C$ has $[1:0:0]$ as only singularity (cf. Jacobian), hence your surface has the line $V(y,z)\subset \mathbb P^3$ as set of singularities . At every point of that line the tangent space has dimension $3$.
Tricks of the trade
a) You can use the Jacobian also in projective space
b) In an affine or projective space of dimension $n$ a hypersurface has tangent space of dimension $n$ at a singularity and of dimension $n-1$ at all non-singular points.
Edit: Cones
Since Jonathan asks, here is why a homogeneous polynomial $f(x,y,z)$ not involving $w$ defines the cone $C\subset \mathbb P^3$ with vertex $S=[1:0:0:0]$ over the projective curve $V(f)\subset \mathbb P^2_{0:x:y:z}$
A point $R$ on the line joining $S$ to a point $Q=[0:a:b:c]\in V(f)$ has coordinates $[u:va:vb:vc]$ for some $[u:v]\in \mathbb P^1$.
Since $f(R)=f(va,vb,vc)= v^{deg(f)} f(a,b,c)=0$, we see that indeed $R\in C$ and $C$ is the claimed cone.
If you look at the beginning of section 2, Fulton defines an affine variety to be an irreducible algebraic variety, i.e. a closed irreducible subset of $\Bbb{A}^n$ with the Zariski topology.
Now in the case of $\Bbb{A}^1$, the union of two points is not an affine variety. Finite point sets are closed yes, but then $\{1,2\} = \{1\} \cup \{2\}$ and thus is not an affine variety.
Here's how I would do $(3) \implies (1)$. Let $X = V(I)$ for some ideal $I \subseteq k[X_1,\ldots,X_n]$. Suppose we know that $\dim_k \Gamma(X) < \infty$. Then by Corollary 4 of the Nullstellensatz we get that
$$\dim_k \Gamma(X) = \dim_k k[X_1,\ldots,X_n]/\sqrt{ I} < \infty \implies |V\left(\sqrt{ I}\right)| < \infty.$$
But now $V\left(\sqrt{I}\right) = V(I)$ which means that $|X| = |V(I)| < \infty$. A finite set of points is irreducible iff it consists of a point, so that $X$ is a point.
Q.E.D.
Best Answer
I think proposition 2 in ยง1.6 of Fulton's book might help you: Let $F, G$ be polynomials in $k[X, Y]$ with no common factors. Then $V(F, G)$ is a finite set of points. In your case: $V(f, f_x, f_y) \subseteq V(f, f_x)$. If we assume $f_x \neq 0$ than $f, f_x$ have no common factors as $f$ is irreducible. So we can apply the proposition.