if $f(x,y)=\frac{xy^2}{x^2+y^6}$ and $f(0,0)=0$ show that $f$ is ubounded at any neighborhood of (0,0) 
if $f(x,y)=\frac{xy^2}{x^2+y^6}$ and $f(0,0)=0$ show that $f$ is ubounded at any neighborhood of (0,0) but the restritions to straight line in $\mathbb{R}^2$ is cont.

My attempt:
**Proving continuity **
if $f(x,y)=\frac{xy^2}{x^2+y^6}$  then $y=mx+c$ is equation of any line then we can get $f(x,mx+c)= \frac{x.(mx)^2+c^2.x+2mxc}{x^2+(mx+c)^6}$ then the function is continuous at any point $(x,y)$ which is on the line$y=mx+c$,
We pick any point $(a,ma+c)$ on the line $y=mx+c$,then we try to show that the function is continuous at the point
We know that $f(a,ma+c)= \frac{ma^3+c^2.a+2mac}{a^2+(ma+c)^6}$ .
Now, $\lim_{x \to a}(f(x,mx+c))=\lim_{x \to a}\frac{x^3m^2+c^2x+2mxc}{x^2+(mx+c)^6}=f(a,ma+c)$(by the property of limits).
Proving uboundedness
In this part I did some back calculation . I showed that the function will not be bounded if the numerator becomes infinity or when the denominator becomes infinity.
For the denominator to become $0$ we see that $y^6+x^2=0$ then $(y^2)^3= -x^2$ then $y^2=(-x)^{\frac{2}{3}}$ . So I have shown that if $(x,y) \to (0,0) $ along the given curve $y^2=(-x)^{\frac{2}{3}}$ the limit of the function $f$ doesnt exist.
I am not sure whether this is correct but this has been essentially my attempt.
 A: Yes, the path you used in your attempt (as given in the main comments) works.

Below I outline a strategy for discovering such a path . . .

To show that $g$ is unbounded in a punctured neighborhood of $(0,0)$, it suffices to 
find a path terminating at $(0,0)$ such that $g(x,y)$ approaches infinity as $(x,y)$ approaches $(0,0)$ along that path.

Consider a path in the first quadrant with equation $y=x^a$, where $a$ is a positive constant, left unknown for now.

Then for $(x,y)$ on that path with $(x,y)\ne (0,0)$, we have
$$
g(x,y)=\frac{xy^2}{x^2+y^6}=\frac{x^{2a+1}}{x^2+x^{6a}}
$$
so we would like to find $a > 0$ such that
$$
\lim_{x\to 0^+}\frac{x^{2a+1}}{x^2+x^{6a}}=\infty
$$
To get that to happen, we want $a > 0$ to be such that
$$
\left\lbrace
\begin{align*}
&2a+1 < 2\\[4pt]
&2a+1 < 6a\\[4pt]
\end{align*}
\right.
$$
or equivalently, ${\large{\frac{1}{4}}} < a < {\large{\frac{1}{2}}}$.

Thus we can choose $a={\large{\frac{1}{3}}}$ to achieve the desired result.
A: Equation $y^6+x^2=0$ has only one solution in #\mathbb{R}^2$, and that is $(0,0)$.
To prove that the function is unbounded in the neighbourhood of $(0,0)$ consider (for $y\neq 0$)
$$ f(y^3,y) = \frac{y^3 \cdot y^2}{(y^3)^2+ y^6} = \frac{1}{2y}$$
