Differential multi-variable function limit using polar coordinates does not work Given:
$$ f(x,y) = \frac{xy^3}{x^2 + y^6}$$
Is it differentiable at $(0,0)$ ?
I said no, as it is not even continuous by the path $x = y^3$
However, when we go to polar view that is: $x = r \cos(\theta) , y = r \sin(\theta)$
we get that $$\lim_{r \rightarrow 0^+} \frac{r^4 \cos(\theta) \sin^3 (\theta)}{r^2(\cos^2(\theta) + r^4 \sin^6(\theta))}$$  and thus:
$$\lim_{r \rightarrow 0^+} \frac{r^2 \cos(\theta) \sin^3 (\theta)}{(\cos^2(\theta) + r^4 \sin^6(\theta)}$$
And we can just plot $r = 0$ and get that it is indeed continuous...
What is wrong with this way? I don't understand, as we are taught to use this way to prove or disprove continuity every time, but I did not check if it actually worked or not! I assumed this way works every time, so why does it fail here?

The problem is that : $\lim_{r \rightarrow 0^+} \frac{r^2 \cos(\theta) \sin^3 (\theta)}{(\cos^2(\theta) + r^4 \sin^4(\theta)}$  is not always determinate contrary to what you probably think. What happens if $\theta$ assumes a value which makes denominator $0$ as $r\to 0$. You have indeterminate form $(0/0)$. 
As it turns out, particularly in this case, the above limit depends upon $\theta$, i.e., the direction through which you approach $(0,0)$. Try doing $r\to 0$ along $\theta =\pi/2$. See what happens.
