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.
By definition, the rank of a matrix is equal to the dimension of the span of its columns.
The columns of the $n \times n$ matrix $J$ are precisely $\theta(f_1), \dots , \theta(f_t)$, where each $\theta(f_i)$ is viewed as a vector in $k^n$. Therefore, the rank of $J$ is equal to the dimension of $\text{Span}(\theta(f_1), \dots , \theta(f_t)) \subset k^n$.
I claim that $\theta(\mathfrak b) = \text{Span}(\theta(f_1), \dots , \theta(f_t))$.
- Proving $\theta(\mathfrak b) \subset \text{Span}(\theta(f_1), \dots , \theta(f_t)) $: Observe that if $g \in \mathfrak b = (f_1, \dots , f_t)$, then there exist $a_1, \dots, a_t \in k[x_1, \dots , x_t]$ such that $g = a_1f_1 + \dots + a_t f_t$. Then $\frac{\partial g}{\partial x_i} (p) = a_1(p)\frac{\partial f_1}{\partial x_i} (p) + \dots a_t(p)\frac{\partial f_t}{\partial x_i} (p)$, by the "product rule" for differentiation. (Note that the product rule also produces terms like $f_i(p)\frac{\partial a_i}{\partial x_i} (p)$, but these vanish because $f_i(p) = 0$.) Thus $\theta(g) = a_1 \theta(f_1) + \dots a_t \theta(f_t)$. Hence $\theta(g) \in \text{Span}(\theta(f_1), \dots , \theta(f_t))$.
- Proving $\text{Span}(\theta(f_1), \dots , \theta(f_t)) \subset \theta(\mathfrak b) $: If $v \in \text{Span}(\theta(f_1), \dots , \theta(f_t))$, then there exist $c_1, \dots, c_t \in k$ such that $v = c_1 \theta(f_1) + \dots + c_t \theta(f_t)$. Then $v = \theta(h)$, where $h = cf_1 + \dots + cf_t \in k[x_1, \dots, x_n]$.
To summarise, the rank of $J$ is equal to the dimension of $\text{Span}(\theta(f_1), \dots , \theta(f_t))$ as a vector subspace of $k^n$, and $\text{Span}(\theta(f_1), \dots , \theta(f_t)) = \theta(\mathfrak b)$. Hence the rank of $J$ is equal to the dimension of $\theta(\mathfrak b)$ as a vector subspace of $k^n$.
Best Answer
Question: "Is it always true that $dim(V)≤n− rank (J_p(I))$?"
Answer: I use the notation of Hartshorne, Chapter I. If $(A, \mathfrak{m})$ is a noetherian local ring with residue field $k$ it follows (HH.I.Prop.5.2A)
$$krdim(A) \leq dim_k(\mathfrak{m}/\mathfrak{m}^2).$$
There is a formula (see the proof of HH.Thm.I.5.1) saying
$$rk(J_p)=n-dim_k(\mathfrak{m}_p/\mathfrak{m}^2_p)$$
Here we assume $k$ is algebraically closed. Assume $Y:=V(I) \subseteq \mathbb{A}^n_k$ (in the notation of chapter I, HH) is an algebraic variety with coordinate ring $A(Y):=k[x_1,..,x_n]/I$ and maximal ideal $\mathfrak{m}_p$ corresponding to the point $p\in Y$. Localize $A:=A(Y)$ at $\mathfrak{m}:=\mathfrak{m}_p$ you get a noetherian local ring $(A_{\mathfrak{m}}, \tilde{\mathfrak{m}})$ with
$$krdim(A)=krdim(A_{\mathfrak{m}}) \leq dim_k(\tilde{\mathfrak{m}}/\tilde{\mathfrak{m}}^2)=dim_k(\mathfrak{m}/\mathfrak{m}^2).$$
Hence
$$dim(V):=krdim(A) \leq dim_k(\mathfrak{m}/\mathfrak{m}^2) =n-rk(J_p(I))=dim_k(T_p(V)).$$
Edit: Let me briefly explain why I think this is true. If $T_p(V)$ is the tangent space of $V$ at $p$, then I think the following two statements are true: $dim(V)≤dim_k T_p(V)$ and $dim_k(T_p(V))=n−rank(J_p(A))$. They immediately imply what I stated. Am I correct?
Answer: By the above argument: This claim is correct.