Determining the tangent space of a submanifold of $\Bbb C^3$

Question

Define a map $f:\Bbb C^3\to \Bbb C^2$ by $f(x,y,z)=(x^2-y,xy-z)$ and let $S$ be the level set $f^{-1}(0)$. It is easily shown that $0$ is a regular value, so $S$ is an embedded complex curve in $\Bbb C^3$. How can we show that the tangent space $T_pS$ where $p=(0,0,0)$ contains the tangent vector $\dfrac{\partial}{\partial x}$? I know that there is a characterization $T_pS=\ker(df_p)$, so I tried to show that $\dfrac{\partial}{\partial x}\in \ker(df_p)$, which is equivalent to $0=(df_p)\left(\dfrac{\partial}{\partial x}\right)g=\dfrac{\partial}{\partial x}|_{p}(g\circ f)$ for all smooth $g:\Bbb C^3\to \Bbb C$, but I got stuck. Is there another approach? Thanks in advance.

Answer 1

If we explicitly compute the differential map as a matrix in the basis $\{\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\}$, we see that $$df_p = \begin{bmatrix} \frac{\partial (x^2 - y)}{\partial x} & \frac{\partial (x^2 - y)}{\partial y} & \frac{\partial (x^2 - y)}{\partial z}\\ \frac{\partial (xy - z)}{\partial x} & \frac{\partial (xy - z)}{\partial y} & \frac{\partial (xy - z)}{\partial z} \end{bmatrix} = \begin{bmatrix} 2x & -1 & 0\\ y & x & -1 \end{bmatrix}.$$ When evaluating the differential at the point $p = (0,0,0)$, our linear map from above reduces to $$ df_p= \begin{bmatrix} 0 & -1 & 0\\ 0 & 0 & -1 \end{bmatrix}.$$ The first column corresponds to the vectors in the kernel of the differential map, and these are precisely the vectors that belong in the span of $\frac{\partial}{\partial x}$, which proves the claim.