Is $f(x,y)=\frac{xy^3}{x^2+y^6}$ differentiable at $(0,0)$? Is the following function differentiable at $(0,0)$?
$$ \
 f(x,y) =
  \begin{cases}
   \frac{xy^3}{x^2+y^6} & \text{if } (x,y) \ne (0,0), \\
   0       & \text{if } (x,y) = (0,0).
  \end{cases}
$$
I found that both of the partial derivatives are $0$, and then tried to calculate the following limit:
$$\lim_{(x,y) \to (0,0)} \frac{\frac{xy^3}{x^2+y^6}}{\sqrt{x^2+y^2}} = \lim_{(x,y) \to (0,0)} \frac{xy^3}{(x^2+y^6) \sqrt{x^2+y^2}}$$
And then I got stuck. I tried the squeeze theorem, but I still couldn't calculate it.
How can I calculate this limit?

Recall that continuity is a necessary condition for differentiability since differentiability implies continuity and by $y^3=v \to 0$ using polar coordinates we have
$$\frac{xy^3}{x^2+y^6}=\frac{xv}{x^2+v^2}=\cos\theta\sin \theta$$
