differentiability at a point (0,0) based on partial derivatives

Question

For $$ f(x,y)=\begin{cases} y^2 sin\left(\frac{x}{y}\right) & \text{if } y\neq0 \\ 0 & \text{if } y=0 \end{cases}$$ i've shown that it is continuous and that the partial derivatives exist in $\mathbb{R}^2$. However, it appears that the partials are not continuous at $(0,0)$; is this sufficient to show that $f$ is not differentiable at $(0,0)$?

Answer 1

No it is not. A function is continuously differentiable on an open set iff it is partially continuously differential on that set. But there are certainly functions, which are differentiable, but not continuously differentiable.

You have to show that the partial derivatives do not coincide with a common linear map applied to the direction of the derivation.

Answer 2

Absolutely not and here's a good counterexample. To prove that a function is continuous but not differentiable, you have to demonstrate there is no linear mapping that satisfies the limit operation in $R^3$. This can be tricky, as it involves showing this limit doesn't exist at at least one point and this can be true even if the partial derivatives exist.