Let $(V,0)\subset (\mathbb{C}^n,0)$, $n\geq 3$ a (germ of) hypersurface given by $V =\{f=0\}$, $f$ a germ of holomorphic function. A (germ of) holomorphic vector field $X$ on $(\mathbb{C}^n,0)$ is "tangent to $V$" if $X(f) = {\rm d}f(X)$ belongs to the ideal generated by $f$ in $\mathcal{O}\_{\mathbb{C}^n,0}$. 


My question is: **For any hypersurface germ $V$, does there exist a germ of vector field with an isolated singularity that is tangent to $V$?**


If $V$ is smooth or has an isolated singularity, this is fairly simple. However, for germs of non-isolated hypersurface singularities I could neither prove this nor find a counterexample.



Holomorphic vector fields tangent to a hypersuface singularity
I think that the answer to this question is no for $\mathbb C^3$, for 
$$f=zy(z-y)(z-xy).$$
I'll assume that $v$ is holomorphic and tangent to $f=0$ in a small neighbourhood of $(0,0,0)$.


*Proof.* Suppose by contradiction that $v$ is a holomorphic vector field tangent to $f=0$ near $(0,0,0)$ with an isolated zero at $(0,0,0)$. Note that the surface $f=0$ has a singularity along the $x$-axes. So $v$ should be tangent to the $x$-axes and also should be non-zero on it by our assumptions. I claim that this is not possible.


Indeed, note that the tangent cone of the surface $f=0$ at a point $(x\_0,0,0)$ is the union of $4$ planes, $z=0$, $y=0$, $z-y=0$ and $z-x\_0y=0$. These planes have double ratio $x\_0$ (to define such a double ratio intersect these $4$ planes with any plane passing through $(x\_0,0,0)$ and take the double ratio of $4$ lines in the intersection). On the other hand if $v$ were non-zero on the $x$-axis, its flow would send points of the $x$-axes to different points and so it would not preserve the double ratio. This is a contradiction, since a byholomorphism must preserve such a ratio. QED.


