Determine if the intersection of two surfaces is a non-singular curve Let $S_1$, $S_2$ be two surfaces in $\mathbb{R}^3$ implicitly defined by, respectively, $f_1(x,y,z)=0$ and $f_2(x,y,z)=0$. How do I check if the curve $S_1 \cap S_2$ has singular points without explicitly deriving a parametrization? Example:
$$S_1 = \{x^3 + 3(y^2+z^2)^2=2\} \\ S_2 = \{x^4 = y^3\}$$
I see no obvious way for a parametrization of $S_1 \cap S_2$, so how do I go about checking if it has singular points?
And, as a bonus more general question: in situations like this, how much information about the intersection curve can I derive just by looking at the surfaces?
 A: Recall that $\nabla f_2(x,y,z)=\left(\frac{
\partial}{\partial x}f_1(x,y,z),
\frac{\partial}{\partial y}f_1(x,y,z),
\frac{\partial}{\partial z}f_1(x,y,z) 
\right)$ is normal to $S_1$ and $\nabla f_2(x,y,z)=\left(\frac{
\partial}{\partial x}f_2(x,y,z),
\frac{\partial}{\partial y}f_2(x,y,z),
\frac{\partial}{\partial z}f_2(x,y,z) 
\right) $ is normal to $S_2$. The vectorial product $\nabla f_1(x,y,z)\times \nabla f_1(x,y,z)$  is a multiple of the vector tangent to the curve $S_1\cap S_2$.
A: Write down the Jacobian matrix 
$$
\big{(}\frac{\partial f_i}{\partial x_j}\big{)}.
$$
A point will be singular iff the Jacobian matrix doesn't have full rank.
In this case, the matrix is
$$
\begin{pmatrix}
3x^2 & 6y(y^2+z^2) & 6z(y^2+z^2) \\
4x^3 & -3y^2 & 0
\end{pmatrix}.
$$
Let's say $x = 0$. Then this matrix will fail to have full rank if $18y^2z(y^2+z^2) = 0$, i.e. if $y$ or $z$ is $0$. Staring at the equations for a minute, we see that  the two points $(0, 0, \pm \sqrt[4]{\frac{2}{3}}$) will be singular points.
If $x \neq 0$, then column 3 is linearly independent of column $1$ unless $z = 0$. In that case, column 2 is $\begin{pmatrix} 6y^3 \\ -3y^2 \end{pmatrix}$, and so there could be a singular point where $-9x^2 -24x^3y^3 = 0,$ i.e. where $3+8xy^3 = 0$. Such a point is on $S_2$ iff $3 + 8x^5 = 0$, i.e. $x = \sqrt[5]{-\frac{3}{8}}$. Now we can solve for $y$ and see that such a point isn't on $S_1$, so there's no other singular points beyond the two we already found.
