Q:

No common factors implies functional independence

Let $f$ and $g$ be two polynomials in two real variables $x$ and $y$, both of which vanish at $(0,0)$. Suppose that they have no common factors. Is it true that their Jacobian
$$J(f,g)=\begin{bmatrix}\partial_x f & \partial_y f\\\partial_x g & \partial_y g\end{bmatrix}$$ is invertible at $(0,0)$?
Typical example would be $f(x,y)=x$ and $g(x,y)=y$.
A one-dimensional analogue seems to be that if two polynomials have no common factors, they are linearly independent as functions.

A:

No, that's not true. You can take $f=y-x^2$ and $g=y$, then 
$$
J(f,g)=\left[\begin{matrix} -2x & 1 \\ 0 & 1 \end{matrix}\right]
$$
which is not invertible at $(0,0)$. The reason is that  the parabola cut out by $f$ and the line cut out by $g$ intersect doubly at the origin, so the resulting point is not a smooth point. 

