Find appropriate function to get a bounded curve Is it possible to choose $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that
$$
a+yf(x,y) = 0, \;\;\;a>0
$$
is a bounded curve?
EDIT: I am very sorry for not being precise enough. I've been searching for an answer for some time now and I've come to think of some things for granted... Anyway:
ADDENDUM: $f$ should be continously differentiable with respect to both its variables. The trivial $f(x,y)=0$ is not accepted.
 A: There is no single continuous $f:\ {\mathbb R}^2\to{\mathbb R}$ that produces a nonempty bounded curve for all $a>0$.
Proof. Let $g(x,y):=y\ f(x,y)$. Then $\inf g(x,y)=-\infty$, or else the solution set $$S_a:=\{(x,y)\ |\ a + y\ f(x,y)=0\}$$ would be empty for large enough $a$. Therefore for  arbitrary $R>a$  one can find a point $(\xi,\eta)\in{\mathbb R}^2$ with $\xi^2+\eta^2>R^2$ and $g(\xi,\eta)<-R$. Connect the point $(R,0)$ with $(\xi,\eta)$ by a curve $\gamma:\ t\to z(t)$ $\ (0\leq t\leq 1)$ such that $|z(t)|\geq R$ for all $t\in[0,1]$. As $g(R,0)=0$ and $g(\xi,\eta)<-R$ there has to be a $\tau\in[0,1]$ with $g\bigl(z(\tau)\bigr)=-a$. It follows that $z(\tau)\in S_a$ and at the same time $|z(\tau)|\geq R$. As $R>a$ was arbitrary, $S_a$ is unbounded.$\quad {}_\square$
When $f$ is allowed to depend on $a$ (or $a$ is bounded away from $\infty$) then one can construct examples: For a fixed $b>0$ consider the function
$$f(x,y):={2b\over 1+x^2+y^2}\ .$$
I claim that the solution set $S_a$ is a circle for any $a$ with $0<a<b$.
Proof.  Since $1+x^2+y^2>0$ for all $(x,y)$ the equation $a+y f(x,y)=0$ is equivalent with
$$(1+x^2+y^2) + 2y{b\over a}=0$$
or
$$x^2 +\Bigl(y+{b\over a}\Bigr)^2={b^2-a^2\over a^2}\ .$$
When $0<a<b$ this is the equation of a circle in the $(x,y)$-plane. $\quad{}_\square$
A: Seems like $f(x,y)=\frac{1}{y}$ (except $f(x,0)=0)$ should do the trick $\ldots$
A: What are the most basic types of bounded curves we know? Ellipses.
If you choose $f(x,y) = \frac{y}{x^2-1}$ it should work.
Define the value of $f(1,y)$ to be whatever you want. This way you do get a curve, not the empty set.
This looks a lot like homework.
Edit: sign mistake
A: For $1 \leq x \leq 2$ define $f(x, y) = -a$. For other values of $x$ define $f(x, y) = 0$.
The resulting set is an interval in $\mathbb{R}^2$, between the points $(1, 1)$ and $(2, 1)$.
