I leanred, if all partial derivatives exist and all are continuous, then it is differentiable. Am I wrong?

I tried same way for this problem, I think it is differentiable because

all the derivative exist and are continuous.

However, it is not differentiable at 0.

Why is it the case?

![The problem](https://i.stack.imgur.com/UWjHD.jpg)

![Another problem](https://i.stack.imgur.com/ozRR1.jpg)

// then I think... the functions of tho problems all have discontinuous partial derivative at (0,0)... right?

then, why is it differentiable or not?..
['https://math.stackexchange.com/questions/425937', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/78027/']
The partial derivatives evaluate to both to $0$. Thus, **if** the function is differentiable, it must be the case the differential is identically zero.

You're looking at $$\lim\_{(x,y)\to (0,0)}\left|\frac{f(x,y)-\nabla f(x,y)\cdot (x-y) }{\lVert (x,y)\rVert}\right|$$

This is $$\lim\_{(x,y)\to (0,0)}\left|\frac{xy}{x^2+y^2}\right|$$

What happens if you take the limit along $y=x$? Is it $0$, as it should be?

**ADD** The partial derivatives are $$\eqalign{
 & \frac{{\partial f}}{{\partial x}} = \frac{{{y^2}}}{{{x^2} + {y^2}}}\frac{y}{{\sqrt {{x^2} + {y^2}} }} \cr 
 & \frac{{\partial f}}{{\partial y}} = \frac{{{x^2}}}{{{x^2} + {y^2}}}\frac{x}{{\sqrt {{x^2} + {y^2}} }} \cr} $$

Are you **sure** they are continuous at the origin?
For the first one.
if $y = ax$,
$\dfrac{xy}{\sqrt{x^2+y^2}}
=\dfrac{ax^2}{\sqrt{x^2+a^2x^2}}
=\dfrac{a}{\sqrt{1+a^2}}
$,
so the function is not even continuous
at $(0, 0)$.

(This is, of course, not original.)