Find $f$ such that $f^{-1}(\lbrace0\rbrace)$ is this knotted curve (M.W.Hirsh)

An alternate, algebraic approach: Let $S^3$ be the sphere $|z_1|^2+|z_2|^2=2$ in $\mathbb{C}^2$. Then the trefoil knot $K$ is given by the equation $z_1^3 = z_2^2$ in $S^3$, and can be parametrized as $(e^{2 i \theta}, e^{3 i \theta})$. Define $\phi: S^3 \to \mathbb{C}$ by $\phi(z_1, z_2) = z_1^3 - z_2^2$. So $K = \phi^{-1}(0)$.

I claim that $\phi$ is a submersion along $K$. If we consider $\phi$ as a map $\mathbb{C}^2 \to \mathbb{C}$, it is a submersion everywhere except at $(0,0)$. To verify that it is still a submersion when restricted to $S^3$, we must check that the $2$-dimensional kernel of $D \phi$ is transverse to the $3$-dimensional tangent space to $S^3$ at every point of $K$. To do this, I just have to give an element of $\mathrm{Ker}(D \phi)$ which is not in $TS^3$. Namely, at the point $(e^{2 i \theta}, e^{3 i \theta})$, the vector $(2 e^{2 i \theta}, 3 e^{3 i \theta})$ is in $\mathrm{Ker}(D \phi)$ but not in $TS^3$. (We can think of this vector as the derivative of $(e^{2 (t+i \theta)}, e^{3 (t+i \theta)})$ with respect to $t$. This curve lies on $z_1^3=z_2^2$, so its derivative is in $\mathrm{Ker}(D \phi)$, but it is clearly transverse to $S^3$.)

So we have a $C^{\infty}$ map from $S^3$ to $\mathbb{C}$ where $\phi^{-1}(0)=K$ and $\phi$ is a submersion along $K$. To convert to a map $\mathbb{R}^3 \longrightarrow \mathbb{R}^2$, just remove a point from $S^3$: Remove a point on $K$ to make a knot which stretches off to infinity as in the diagram, or remove a point not on $K$ to make a closed knot.

Here is a picture of your knot, in stereographic projection from the point $(1,1) \in S^3$:

It can be given parametrically as $$\left( \frac{\cos (2 t)-\cos (3 t)}{-\cos (2 t)-\cos (3 t)+2}, \ \frac{\sin (2 t)}{-\cos (2 t)-\cos (3 t)+2},\ \frac{\sin (3 t)}{-\cos (2 t)-\cos (3 t)+2}\right )$$

I get that the explicit coordinates of $\phi(u,v,w)$ are given by $$\begin{multline} \frac{1}{(1 + u^2 + 2 v^2 + 2 w^2)^3} \\ \left(-2 + 2 u - 12 u^2 - 4 u^3 + 6 u^4 + 10 u^5 + 56 v^2 - 120 u v^2 - 40 u^2 v^2 + 40 u^3 v^2 - 104 v^4 + 40 u v^4 + 24 w^2 - 24 u w^2 + 24 u^2 w^2 + 40 u^3 w^2 - 80 v^2 w^2 + 80 u v^2 w^2 + 24 w^4 + 40 u w^4, \right. \\ \left. 12 v - 48 u v + 24 u^2 v + 48 u^3 v + 12 u^4 v - 112 v^3 + 96 u v^3 + 48 u^2 v^3 + 48 v^5 + 8 w + 16 u w + 16 u^3 w - 8 u^4 w + 32 u v^2 w - 32 u^2 v^2 w - 32 v^4 w - 48 v w^2 + 96 u v w^2 + 48 u^2 v w^2 + 96 v^3 w^2 + 32 u w^3 - 32 u^2 w^3 - 64 v^2 w^3 + 48 v w^4 - 32 w^5 \right) \end{multline}$$ I got this by composing the inverse of sterographic projection, $$\frac{1}{1+u^2+2 v^2+2 w^2} \left(u^2+2 u+2 v^2+2 w^2-1,4 v,u^2-2 u+2 v^2+2 w^2-1,4 w\right)$$ and the map $z_1^3 - z_2^2$.

Main construction

Let $\gamma$ be the trefoil knot in $\mathbb{S}^3$. Since $\gamma$ is a fibered knot (see D. Rolfsen, Knots and Links, Chapters H and I), there is a tubular neighborhood $N\gamma \simeq \mathbb{S}^1\times\mathbb{D}^2$ and a fibration $f': \mathbb{S}^3\backslash\langle\gamma\rangle \rightarrow \mathbb{S}^{1}$ such that $$ f'(x,y) = \frac{y}{|y|}\quad\text{for all } (x,y)\in \mathbb{S}^1\times (\mathbb{D}^2\backslash\{0\}). $$ The projection to the second factor of $\mathbb{S}^1\times\mathbb{D}^2$ determines a smooth map $f'': N\gamma \rightarrow \mathbb{D}^2$ with regular value $0$ and with $(f'')^{-1}(0)=\langle\gamma\rangle$. We define $f: \mathbb{S}^3 \rightarrow \mathbb{D}^2$ by $$f(z):=\begin{cases} f''(z) &\text{for }z\in N\gamma, \text{ and by}\\ f'(z) & \text{otherwise}. \end{cases}$$ The compatibility condition guarantees that $f$ is continuous. Notice that the entire complement of $N\gamma$ gets mapped into $\mathbb{S}^1 = \partial \mathbb{D}^2$.

Some details

1) Looking at the figure, we extend the curve through the boundary points outside of $\mathbb{D}^3$ and connect the two ends there to obtain the trefoil $\gamma: \mathbb{S}^1 \rightarrow \mathbb{R}^3$. We then pick a smooth embedding $\psi: \mathbb{R}^3\rightarrow \mathbb{S}^3$, transfer everything to $\mathbb{S}^3$, and construct $f: \mathbb{S}^3 \rightarrow \mathbb{D}^2$. In the end, we consider the restriction $f\circ \psi: \mathbb{D}^3 \rightarrow \mathbb{D}^2$.

2) The main construction gives a continuous extension $f: \mathbb{S}^3 \rightarrow \mathbb{D}^2 \subset \mathbb{R}^2$ of the smooth map $f'': N\gamma \rightarrow \mathbb{D}^2$ such that $f(\mathbb{S}^3 \backslash N\gamma)\subset \mathbb{S}^1$. To make $f$ smooth, we apply Theorem 2.5 from Chapter I of A. Kosinski, Differentiable manifolds:

Theorem 2.5: Let $f:M\rightarrow \mathbb{R}^n$ be a continuous map, smooth on a closed subset $K$ of $M$, and let $\varepsilon>0$. Then there is a smooth map $g: M \rightarrow \mathbb{R}^n$ that agrees with $f$ on $K$ and such that $|f(p) - g(p)| < \varepsilon$ for all $p\in M$.

Applying this theorem to $f$, $M=\mathbb{S}^3$, $K=\frac{3}{4}N\gamma$, $n=2$ and $\varepsilon = \frac{1}{2}$, we obtain a smooth map $g: \mathbb{S}^3 \rightarrow \mathbb{R}^2$ which equals $f$ on $K$ and such that $g^{-1}(0) = f^{-1}(0)$. Since $\mathbb{S}^3$ is compact, we can scale $g$ by a constant to achieve $g(\mathbb{S}^3)\subset \mathbb{D}^2$. The scaling changes neither the regularity of $0$ nor the level set $g^{-1}(0)$. This $g$, after the procedures from 1), is then the map $f$ we have been looking for.

Acknowledgment

The problem was solved with the help of Prof. Dr. Urs Frauenfelder who suggested to look in the Rolfsen's book for the precise definition of a fibered knot.

Reaction to comments:

@Moishe Kohan: It seems like that you are using a more general definition of a fibered knot where the compatibility condition does not necessarily hold, e.g., the definition from J. Harer, How to construct all fibered knots. This definition requires only that the complement $W:=M\backslash K$, where $M$ is a $3$-manifold and $K$ the knot, is a fibration over $\mathbb{S}^1$ such that closures of its fibers $F$ are Seifert surfaces. We then have $W\simeq F \times [0,1]/\sim$, where the ends are identified by a general homeomorphism $h$ of $F$. Then, of course, the intersection $F \cap N\gamma$, which is a knot itself, might link with $K$ non-trivially, depending on $h$.

However, the definition of a fibred knot from the Rolfsen's book, which I am using, imposes the compatibility condition, and hence link $link(F\cap N\gamma, K)=0$. This is a screenshot from the Rolfsen's book:

A fibration of the trefoil satisfying this definition is constructed explicitly in Chapter I of the book.

According to your terminology, a fibration satisfying $link(F\cap N\gamma, K)=0$ is called a Seifert fibration. Hence, it seems like that Rolfsen defines and works with Seifert fibrations implicitly.

To complete the reply to your comment, I sum up the issue of the regular value here:

It holds $f^{-1}(0)=(f'')^{-1}(0)=\langle\gamma\rangle$ because the image of $f'$ lies inside of $\mathbb{S}^1$. Also, $0$ is a regular value of $f''$ since $f''$ is just a projection to the second component. It is also a regular value of $f$ as it agrees with $f''$ on a neighborhood of $\langle\gamma\rangle$. The smoothening of $f$ does not change these facts.