Restriction of a polynomial to an approximated algebraic curve

We are given two polynomials in the plane $$f(x,y)$$ and $$g(x,y)$$. Both define an algebraic curve in the plane.

Let $$\hat{h}(x)$$ be the Taylor expansion of the function $$h$$ at a point $$(x_0,y_0)$$ defined by $$f(x,h(x))=0$$ (computed by Implicit Function Theorem).

Is it true that, there exists a neighbourhood of $$(x_0,y_0)$$ such that $$sign(g(x,y)_{\mid f(x,y)=0})=sign(g(x,y)_{\mid y=\hat{h}(x)})$$

?

We have: $$h(x)=h(x_0)+(x-x_0)h'(x_0)+o(x-x_0)$$ and $$\hat h(x)=h(x_0)+(x-x_0)h'(x_0)$$

then $$g(x,y)_{\mid f(x,y)=0}=g(x,h(x))=g(x,y_0) + \frac{\partial g}{\partial y}(x_0,y_0)((x-x_0)h'(x_0)+o(x-x_0))$$

On the other hand, $$g(x,y)_{\mid y=\hat h(x)}=g(x,\hat h(x))=g(x,y_0) + \frac{\partial g}{\partial y}(x_0,y_0)((x-x_0)h'(x_0)$$ so my statement would be obvious ?

Assuming $$g(x_0,y_0)\neq 0$$, this is trivial: by continuity of $$g$$, there is a neighborhood $$U$$ of $$(x_0,y_0)$$ in the plane on which $$g$$ has constant sign. In particular, inside that neighborhood it has the same sign whether you restrict it to the vanishing set of $$f$$ or to the graph of $$\hat{h}$$.
If $$g(x_0,y_0)=0$$, there is no reason for something like this to be true. For instance, if $$g=f$$, then $$g$$ is always $$0$$ on the vanishing set of $$f$$, but will not be $$0$$ on the graph of $$\hat{h}$$ unless $$\hat{h}$$ happens to actually be exactly equal to $$h$$.