$\lim_{ (x,y)\to(0,0)} f(x, y)$ exists along all parabolas that contain the origin. Give a proof or counterexample of the following statement: Let $f$ be a real-valued
function, that is defined and continuous on all of $\mathbb{R}^2$ except at the origin. It has a removable discontinuity at the origin provided that the limit
$\lim_{ (x,y)\to(0,0)} f(x, y)$
exists along all parabolas that contain the origin.
 A: Here's an example that satisfies the criteria for parabolas of the form $y=ax^2+bx$ or $x=ay^2+by$:
Let $$f(x,y)={\cases{xy^3\over x^2+y^6,&$(x,y)\ne(0,0)$ \cr 0,\phantom{\biggl|}& otherwise}}.$$ 
First we show that the limit as $(x,y)$ approaches the origin along one of the parabolic paths given above is 0:
Along the parabola $y=ax^2+bx$, $a\ne 0$:
$$f(x,y)={x(ax^2+bx)^3\over  x^2+(ax^2+bx)^6} 
\quad\buildrel{x\rightarrow0}\over\longrightarrow\quad 0,
$$
as two applications of L'Hopital's rule will verify (or observe that the dominant term upstairs is $ax^7$ and the dominant term downstairs is $x^2$).
Along the parabola $x=ay^2+by$,  $a\ne 0$:
$$\eqalign{f(x,y)={(ay^2+by)y^3\over (ay^2+by)^2 +y^6} 
&={ay^5+by^4\over a^2y^4+2aby^3+b^2y^2+y^6}\cr
&={ay^3+by^2\over a^2y^2+2aby +b^2 +y^4}\cr

& \buildrel{y\rightarrow0}\over\longrightarrow\quad 0,}$$
as easily seen when $b\ne 0$. For $b=0$, we have
$$
{ay^3+by^2\over a^2y^2+2aby +b^2 +y^4}
={ay^3 \over a^2y^2  +y^4}={ay \over a^2   +y^2}
\quad \buildrel{y\rightarrow0}\over\longrightarrow\quad 0, 

$$
as well.
Now we show that $\lim\limits_{(x,y)\rightarrow(0,0)} f(x,y)$ does not exist (and thus, $f$ is discontinuous at the origin, but the discontinuity is not removable):
Just observe that along the path $x=y^3$:
$$
f(x,y)={y^6\over 2y^6}\quad\buildrel{y\rightarrow0}\over\longrightarrow\quad {1\over2}.
$$

I'm not sure what happens for a general parabola that passes through the origin...

Incidentally, in, Counterexamples in Analysis, by Bernard R. Gelbaum and John M. H. Olmsted, 
page 116, an example is given of a function 
 which   has no limit at $(0,0)$,
but such that for any path of the form $x^m=(y/c)^n$, where $c\ne 0$ and $m,n$ are relatively prime 
positive integers, the limit as $(x,y)$ approaches the origin  along the path is zero. The function with the stated properties is:
$$
f(x,y)=\cases{ {e^{-1/x^2}y\over e^{-2/x^2}+y^2 },& $x\ne0$\cr 0\phantom{\biggl|} ,&$x=0$}.
$$
