Is $f(x,y)=\frac{xy^3}{x^2+y^6}$ differentiable at $(0,0)$?

Question

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?

Answer 1

It's not even continuous at $(0,0)$. Hint: $f(y^3,y)=\dfrac12$ if $y\ne0$.

Answer 2

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

$$\frac{xy^3}{x^2+y^6}=\frac{xv}{x^2+v^2}=\cos\theta\sin \theta$$