*NB: This answer is directed to the questions about the real case, not the complex case, which was already treated by Francesco. Added 5 July 2021: Because of some questions I have received over the years since this was written, I have realized that there are a few places where the logic and flow are not completely transparent, so I have decided to make a few small changes to clarify those points.*

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 (germs 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 is: **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 (i.e., real-analytic germs). 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.

There remains the question of whether the zero locus of an irreducible real-analytic germ $f$ that is singular at the origin could contain a *smooth* hypersurface passing through the origin.
The answer to this is also 'no', but it takes a little work to see this.

The case of a curve is not hard, using some standard facts about resolution of curve singularities: If $f(x,y)$ is a nonzero, real-valued analytic function defined on a neighborhood of the origin in $\mathbb{R}^2$ that is irreducible in the ring of analytic germs at the origin and satisfies $f_x(0,0)=f_y(0,0)=0$, then the locus $f(x,y)=0$ cannot contain a smoothly embedded curve passing through the origin.

A sketch of a proof is as follows: If the origin is not isolated, then $f(z,w)$ is a $\mathbb{C}$-valued analytic function defined on a neighborhood of the origin in $\mathbb{C}^2$ that is also irreducible in this larger ring, and hence there is a neighborhood of the origin in $\mathbb{C}^2$ such that, in this neighborhood, the locus $f(z,w)=0$ can be parametrized by an embedded disc in $\mathbb{C}$ in the form $(z,w) = (a(\tau),b(\tau))$ where $a$ and $b$ are analytic functions of $\tau$ for $|\tau| < 1$ with $a(0)=b(0)=0$. By a (real) rotation, we can assume that $a$ vanishes to a lower order, say $k>1$, than $b$ does. Thus, we can reparametrize in $\tau$ so that $a(\tau) = \tau^k$ for some $k>1$. In particular, the real locus will be parametrized by some curves of the form $\tau = \omega\,t$ where $t$ is real and $\omega^k = \pm 1$. Choosing one such curve and replacing $t$ by $t/\omega$, we can assume that $(a(t),b(t))$ is real for all small real $t$, and that this parametrizes a 'branch' of the real locus that passes through the origin. In particular, the coefficients of $b$ are real, so our curve is parametrized in the form
$$
(x,y) = \bigl(\ t^k,\ b_l t^l + b_{l+1} t^{l+1} + \cdots\ \bigr)
$$
where $l>k$ and, because of the embeddedness property of the disk, the greatest common divisor of $k$ and those $m$ for which $b_m\not=0$ must be $1$. Thus, the curve is expressed in the form
$$
y = b_l\ x^{l/k} + b_{l+1}\ x^{(l+1)/k} + \cdots
$$
where at least one of the exponents in this series is not an integer. It follows that the function on the right hand side of this equation cannot be smooth at $x=0$, even though, since $l>k$, it is $C^1$.

One conclusion of all this is that, if $g$ is a real-valued *smooth* function on a neighborhood of $0\in \mathbb{R}$ such that $g(0)=0$ and such that there exists a nontrivial real-analytic $f$ defined on a neighborhood of $(0,0)$ such that $f\bigl(x,g(x)\bigr)\equiv0$ for $x$ in the domain of $g$, then $g$ must actually be *real-analytic* in a neighborhood of $0$. (Note, however, that there do exist such 'real-analytically constrained' $g$ that, for some $m>0$, are $C^m$ but not $C^{m+1}$ at $x=0$.)

Now, an easy argument shows that this $1$-variable fact implies the corresponding $n$-variable fact: If $g$ is a real-valued *smooth* function on a neighborhood of $0\in \mathbb{R}^n$ such that $g(0)=0$ and such that there exists a nontrivial real-analytic $f$ defined on a neighborhood of $(0,0)\in\mathbb{R}^n\times\mathbb{R}$ such that $f\bigl(x,g(x)\bigr)\equiv0$ for $x$ in the domain of $g$, then $g$ must actually be *real-analytic* in a neighborhood of $0\in\mathbb{R}^n$. (Basically, the hypotheses and the $1$-variable result imply that $g\circ x$ is real-analytic for any real-analytic germ of a curve $x:(\mathbb{R},0)\to(\mathbb{R}^n,0)$, and this easily implies that $g$ itself is real-analytic in a neighborhood of $0\in\mathbb{R}^n$.)

Thus, we have the answer to the question **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 ***smooth* hypersurface near the origin?

The answer is 'no', because *smooth* would imply *real-analytic*, and we have already seen that this cannot happen.