Does not existence of partial derivatives at a point tells us that the function is not differentiable?

Question

This question arose after my calculus test in which the told us: Show that $$f(x,y)=\begin{cases}\frac{2x^3}{x^2 +y^2}&\text{ if }(x,y)\neq(0,0)\\0&\text{ if }(x,y)=(0,0).\end{cases}$$ is not differentiable at $(0,0)$. I showed that the partials do not exist in $(0,0)$ thus the function is not differentiable.

Answer 1

Proof that $f$ isn't differentiable.

If $f$ was differentiable at $(0,0)$, we would have $$\lim\limits_{(x,y) \to (0,0)} \frac{f(x,y) - f(0,0) -2x}{\sqrt{x^2+y^2}} = 0,$$

as $\frac{\partial f}{\partial x}(0,0)=2$ and $\frac{\partial f}{\partial y}(0,0)=0$.

But $$\frac{f(x,y) - f(0,0) -2x}{\sqrt{x^2+y^2}} = \frac{-2xy^2}{\left(x^2+y^2 \right)^{3/2}} = \pm \frac{1}{\sqrt{2}}$$

for $x=y$. Hence $f$ is not differentiable at $(0,0)$.

Answer 2

Yes, you are right. Because if the function $f$ is differentiable at a point $p$ of its domain, then the directional derivatives exist for every direction $v$. And the partial derivatives are a particular case of directional derivatives.

The only problem is that in your specific example, the partial derivatives do exist. In fact $\frac{\partial f}{\partial x}(0,0)=2$ and $\frac{\partial f}{\partial y}(0,0)=0$.