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Let $(x,y) \in U \subset \mathbb{R}^{2}$, and $f:U \mapsto \mathbb{R}$ smooth. $U$ is such that $0 \leq y \leq\epsilon << 1$. In other words, $y$ is small.

Suppose that, by expanding $f(x,y)$ in $y$ about $y = 0$ for any $x$ I get the following series for $f(x,y)$:

$$f(x,y) =\frac{g(x)}{y^{2}} + \frac{h(x)}{y} + i(x)+...$$ for all $(x,y) \in U$, where $g, h, i$ are smooth function of $x$. So essentially we have $f(x,y) = O(y^{-2})$.

Using these series, I verify (say numerically) that $f(x, y^{*}) \neq 0$ for some choice of $y^{*} \neq 0 \in U$ and all $x$.

Does this imply that $f(x,y) \neq 0$ for all $y \neq 0$ by implicit function theorem, since the series expansion above gives $\frac{\partial f}{\partial y}|y^{*} \neq 0$ ?

Correction: Numerically I verify that $g(x) \neq 0$.

Also, I mean to say $g'(x) \neq 0$ rather than $\frac{\partial f}{\partial y}|y^{*} \neq 0$. Essentially I am asking if $g'(x) \neq 0$ is equivalent in some sense $\frac{\partial f}{\partial x} \neq 0 $ which gives that for small $y$ we can continue the curve on which $f(x,y) \neq 0$...

Thanks.

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    You say that $0 \leq y \leq \epsilon$, but according to your expansion $f \to \infty$ when $y \to 0$. Should the inequality be strict? – 2017-01-27
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    I allow $f \to \infty$ @Rastapopoulos – 2017-01-27

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No. Consider $y^* = \epsilon$ and $f(x,y) = -\frac{\epsilon}{2} + x^2 + y$.

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    Just a question here - did I actually state the implicit function theorem properly for my intention? What if I impose $g'(x) \neq 0$ (Essentially asking that $\frac{\partial f}{\partial x} \neq 0$) ? @Hurkyl – 2017-01-27
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    Your example does not have the series expansion I impose @Hurkyl – 2017-01-27
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    @Alex: $g(x) = 0$, $h(x) = 0$, $i(x) = x^2 - \frac{\epsilon}{2}$, and $\ldots = y$. – 2017-01-27
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    Ah, right. What if I impose $g(x) \neq 0$ and also $g'(x) \neq 0$? (I edited my question to take that into account) @Hurkyl – 2017-01-27