I have a formula of the form $f(x)^2 = 0$, where $x \in \mathbb{R}^n$ and $f$ is a Diophantine polynomial. I am wondering if there is a general way to produce a formula of the form $\exists y \in \mathbb{R}^m \, : \, g(x, y) \ne 0$, where $g$ is a Diophantine polynomial, such that the two formulas pick out the same subset of $\mathbb{R}^n$.
I suspect that this is impossible, but I'm not sure how to go about proving that. Any advice?
Since polynomials are continuous, the solution set to $f(x)=0$ is a closed subset of $\mathbb R^n$.
The solutions to $g(x,y)\ne 0$ is an open subset of $\mathbb R^{n+m}$, and projecting this to $\mathbb R^n$ preserves openness.
Therefore your plan is only possible if the solutions to $f(x)=0$ is both open and closed, that is, if it is either $\varnothing$ or all of $\mathbb R^n$. (And in that case it is indeed, trivially, possible).
On the other hand, if you're speaking of Diophantine ponynomials in particular, perhaps you meant to speak about $\mathbb N^n$ and $\mathbb N^m$ instead of $\mathbb R^n$ and $\mathbb R^m$? In that case it is still impossible, though it is a bit more involved to argue.
First, the existentially quantified $y$s don't improve expressivity. We can view $g(x,y)$ as a polynomial in $y$ whose coefficients are polynomials in $x$. Then $\exists y.g(x,y)\ne 0$ exactly for those $x$ such that the value of one of the coefficients polynomials is nonzero. It is trivial that if all of the coefficient polynomials are zero, then no $y$ can make $g(x,y)$ nonzero. On the other hand, if a polynomial in $y$ has any nonzero coefficient, then it cannot be zero at all integer points (the can be proved by induction on the number of variables), so there will be an $y$ that makes $g(x,y)$ nonzero.
Thus, there is a polynomial $h(x)$ such that $\exists y.g(x,y)\ne 0$ exactly when $h(x)\ne 0$. Namely, $h$ is the sum of the squares of the coefficient polynomials.
On the other hand, if we fix all but one of the coordinates of $x$, the solution set for $h(x)\ne 0$ for the remaining coordinate is either empty or cofinite. On the other hand, under the same fixing the solution set for $f(x)=0$ is either all of $\mathbb N$ or finite. So if $f(x)=0 \iff h(x)\ne 0$, then these one-coordinate solution sets must be either $\varnothing$ or $\mathbb N$. Applying this property to one component at a time, we find that $f(x)=0$ either nowhere or everywhere.