I need to find the limit of a function $f(x,y)$ as $(x,y)\rightarrow (0,0)$. The only method I know of is to consider all paths through $(0,0)$ and do polar coordinate substitution to make it into a limit of $r\rightarrow0$. $$\lim_{(x,y)\rightarrow(0,0)}f(x,y)=\frac{x^3-x^2y}{x^2+y^6}$$ So with polar coordinate substitution that becomes: $$\lim_{r\rightarrow0}f(r,\theta)=\frac{r^3\cos^3\theta-r^3\cos^2\theta\sin\theta}{r^2\cos^2\theta+r^6\sin^6\theta} = \frac{r\cos^2\theta(\cos\theta - \sin\theta)}{\cos^2\theta+r^4\sin^6\theta}\stackrel{?}{=}0$$ This gives the expected answer ($0$, from wolfram alpha) except when $\cos\theta=0$, in which case it's a divide by zero again. If just using an arbitrary straight path $y=ax$ it's pretty easy to prove, but in my understanding that doesn't consider all paths, and thus is not a sufficient for determining the limit.
Any help is appreciated!
$$L =\lim_{(x, y)\to(0, 0)}\frac{x^3-x^2y}{x^2+y^6} = \lim_{(x, y)\to(0, 0)}\frac{x^2}{x^2+y^6}(x-y)$$ Notice $\displaystyle\lim_{(x, y)\to(0, 0)} x-y = 0$ and $\displaystyle\left|\frac{x^2}{x^2+y^6}\right|\le 1$, therefore, $L = 0$.