A limit question of 3-variable-functions. $\lim\limits_{ (x,y,z) \to (0,0,0)} \frac {xyz^2}{x^2+y^4+z^6}$ 
$$\lim\limits_{ (x,y,z) \to (0,0,0)} \frac {xyz^2}{x^2+y^4+z^6}$$

I checked that the limit does not exist but I cannot prove that.
I tried $y=mx$, $z=nx$ and also $y=x^m$, $z=x^n$ but they gave me nothing but the limit equals to zero.
Thanks a lot
 A: Let consider
$$\left|\frac {xyz^2}{x^2+y^4+z^6}\right|=\frac {|x||y|z^2}{x^2+y^4+z^6}$$
and let
$$\begin{cases}
|x|=|X|\\\\
|y|=\sqrt{|Y|}\\\\
z=\sqrt[3] Z
\end{cases}$$
then
$$\frac {|x||y|z^2}{x^2+y^4+z^6}=\frac {|X|\sqrt{|Y|}\sqrt[3] {Z^2}}{X^2+Y^2+Z^2}=\frac{\rho^{1+\frac12+\frac23}f(\theta,\phi)}{\rho^2}=\frac{\rho^{\frac{13}6}f(\theta,\phi)}{\rho^2}=\rho^\frac16f(\theta,\phi)\to 0$$
therefore since
$$\frac {|x||y|z^2}{x^2+y^4+z^6}\to 0 \implies \frac {xyz^2}{x^2+y^4+z^6}\to 0$$
A: An always good way is to use spherical polar coordinates
$$\begin{cases}
x = R\sin\theta\cos\phi \\
y = R\sin\theta\sin\phi \\
z = R\cos\phi
\end{cases}
$$
Which turns the limit into
$$\frac{R^4 \sin (\theta ) \cos (\theta ) \sin ^2(\phi ) \cos ^2(\phi )}{R^6 \cos ^6(\phi )+R^4 \sin ^4(\theta ) \sin ^4(\phi )+R^2 \cos ^2(\theta ) \sin ^2(\phi )}$$
Which goes to zero as $R\to 0$.
But following a linear path like 
$$y = x^n$$
$$z = x^m$$
We get:
$$\frac{x^{2 m+n+1}}{x^{6 m}+x^{4 n}+x^2}$$
The limit of which does not exist.
A: If $x=R\cos\theta,y^2=R\sin\theta\cos\phi,z^3=R\sin\theta\sin\phi$, then the expression is less than $R^{1/6}$.
