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

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 $(0,0)$?

• Isn't. Partial derivatives can be discontinuous while the function is differentiable. – Martín-Blas Pérez Pinilla Feb 16 '15 at 9:25

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.