In some sense, the reason you are running into these 'problems' is that you are working with the ring of real polynomials rather than the ring of real-valued analytic functions, which is also a UFD.  For example, it is not hard to show that there is a (unique) real-valued, real-analytic function $f$ defined in a neighborhood of $0$ and satisfying $f(0)=\frac12$ such that
$$
y^3 + 2x^2y-x^4 = \bigl(y - x^2f(x^2)\bigr)\bigl(y^2 + x^2f(x^2) y + x^2/f(x^2)\bigr),
$$
so $y^3 + 2x^2y-x^4$ is reducible in the ring of real-valued, real analytic functions defined on a neighborhood of the origin.  The curve $y = x^2f(x^2)$ is smooth (in fact, real-analytic, of course), but the $y$-discriminant of the quadratic factor is $x^4f(x^2)^2-4x^2/f(x^2) = -8x^2 + \cdots$, so the only real point of $y^2 + x^2f(x^2) y + x^2/f(x^2)=0$ near the origin is the origin itself.  (The quadratic factor is irreducible in the ring of real-valued analytic functions defined on a neighborhood of the origin.)

Thus, a more easily approached question might be:  **Suppose that the origin is a singular zero of an *irreducible* element $f$ in the ring of real-valued analytic functions defined on a neighborhood of the origin.  Can the zero locus of $f$ be a nonsingular real-analytic hypersurface near the origin?**

The answer to this question is 'no' because near a nonsingular point of such a hypersurface, there will always be a real-analytic function $g$ that vanishes on it whose differential at that point is nonvanishing.  However, this will imply that $g$ is a factor of $f$, which is assumed to be irreducible.