<s>[INST] 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.
 [/INST] 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 $y\neq 0$)
$$ f(y^3,y) = \frac{y^3 \cdot y^2}{(y^3)^2+ y^6} = \frac{1}{2y}$$
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