Singularity of a curve in higher dimensions

Question

Although the problem I am working on is regarding a curve in 6 dimensional affine space, for simplicity, let's consider the Viviani's curve in $\mathbb{C}^3$: $$\mathcal{I}=\langle {x}^{2}+{y}^{2}+{z}^{2}-4, \left( x-1 \right) ^{2}+{y}^{2}-1 \rangle,$$ whose variety $V(\mathcal{I})$ gives the Viviani's curve that has only one double point (a branching) at $(2,0,0)$. The $2 \times 3$ Jacobian is of rank 2 (and hence the kernel is of dimension 1) everywhere except at this point, implying that there is one tangent at every regular point of the curve. Now, I added a point $(0,0,2)$ to $V(\mathcal{I})$ by multiplying $\mathcal{I}$ and $\langle x,y,z-2\rangle$ which yields: $$\mathcal{J}=\langle \left( \left( x-1 \right) ^{2}+{y}^{2}-1 \right) x, \left( \left( x-1 \right) ^{2}+{y}^{2}-1 \right) y, \left( \left( x-1 \right) ^{2}+ {y}^{2}-1 \right) \left( z-2 \right) , \left( {x}^{2}+{y}^{2}+{z}^{2} -4 \right) x, \left( {x}^{2}+{y}^{2}+{z}^{2}-4 \right) y, \left( {x}^{ 2}+{y}^{2}+{z}^{2}-4 \right) \left( z-2 \right) \rangle.$$ By construction ($\sqrt{\mathcal{J}}=\mathcal{I}$), the point $(0,0,2)$ is singular due to multiplicity. My question is how can I prove it just by looking at the ideal $\mathcal{J}$? If I try to calculate the Jacobian for generators of $\mathcal{J}$, at $(0,0,2)$, all its elements are zero and the kernel is $\mathbb{C}^3$ itself! As for planar curves, when there is a multiple point (nodes, cusps or repeated points), the curve has the same number of branches as the multiplicity (real or complex) at that point when the Puiseux series is calculated. But when I calculate the Puiseux series at $(0,0,2)$ for $\mathcal{J}$ (using Singular and Gfan), I get the same thing as for $\mathcal{I}$. Is there any way to determine the correct tangent space of the curve at a singular point of this sort? Like the planar case, am I right to expect additional complex branchings of the curve at this kind of point?