My main question for which I will give an example right below is whether for a partial derivative to exist at a point (say $\frac{\partial f}{\partial x}$) it is necessary for it to be continuous at said point or if it needs only to exist when approaching from the $x$ axis. An example:

Consider the function $$f(x,y)=\frac{x^3+y^3}{x^2+y^2} \quad\forall\vec{x}\in\mathbb{R^2-\vec{0}},\quad f(\vec{0})=0$$ It's obvious it is continuous at $\vec{0}$ since a third degree polynomial goes faster to $0$ than the inverse of a second degree polynomial goes to infinity.

Lets study it's differentiability. For any point except $\vec{0}$ we have

$$ \frac{\partial f}{\partial x}=\frac{(x^2+y^2)3x^2-(x^3+y^3)2x}{(x^2+y^2)^2}$$

$$ \frac{\partial f}{\partial y}=\frac{(x^2+y^2)3y^2-(x^3+y^3)2y}{(x^2+y^2)^2}$$

Naturally the derivatives are not defined for $(x,y)=(0,0)$. If we want to find the derivative with respect to $x$ at $0$ we might try taking the limit to $\vec{0}$ of $ \frac{\partial f}{\partial x}$. We change to polar coordinates ($x=r\cos\theta,y=r\sin\theta$) to see whether the limit exists (if it's unique or path-dependend). After simplyfying we get $$ \lim_{r\rightarrow 0^+}\frac{\partial f}{\partial x}(r,\theta)=\cos^4\theta+3\cos^2\theta\sin^2\theta-2\sin^3\theta\cos\theta $$ We see then that it is path dependent since the value of the limit depends on $\theta$, therefore the limit doesn't exist.

If we don't look at the absolute limit but instead we come from the direction with respect to which we're taking the derivatives, then said limits exist:

$$\lim_{(x,0)\to(0,0)}\frac{\partial f}{\partial x}=1$$ $$\lim_{(0,y)\to(0,0)}\frac{\partial f}{\partial y}=1$$

So my question is, is $Df(0,0)=\begin{bmatrix}1&&1\end{bmatrix}$ the derivative of $f$ at $0$ even though the partial derivatives aren't the same from every direction, or is it that the derivative of $f$ doesn't exist at $(0,0)$?

Thank you for your time.

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    $\begingroup$ $Df$ isn't continuous around the origin, thus it doesn't exist there. You need $$ \lim_{(x,y) \to (0,0)} \partial_x f \quad \text{and} \quad \lim_{(x,y) \to (0,0)} \partial_y f $$ to exist. $\endgroup$ – Jeb Jun 15 '15 at 19:49
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    $\begingroup$ How do you know it isn't continuous around the origin? Certainly the partial derivatives aren't, but the continuity of the partial derivatives isn't necessary for the existence of the total derivative. $\endgroup$ – Cristian Em. Jun 15 '15 at 20:35
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    $\begingroup$ $Df = ( \partial_x f , \partial_y f)$, You've shown that the limits do not exist already... In general, if you want the derivative to exist, the limit has to exist. $\endgroup$ – Jeb Jun 15 '15 at 20:38
  • $\begingroup$ @Jeb: No, Cristian is right, $C^1$ is a stronger condition than differentiability. $\endgroup$ – Hans Lundmark Jun 15 '15 at 21:13

The questions in your first sentence and your last sentence aren't quite the same, but never mind...

The partial derivatives $\partial f/\partial x$ and $\partial f/\partial y$ exist at the origin $(0,0)$, not because of some limit of their values along the axes or otherwise, but directly from the definition of partial derivative: $$ \frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0} \frac{f(h,0)-f(0,0)}{h} = \lim_{h\to 0}\frac{h^3/h^2 - 0}{h} = 1 , $$ and similarly for the other one.

To check whether $f$ is differentiable at the origin, use the definition of differentiability: $$ \lim_{(h,k)\to (0,0)} \frac{f(h,k) - f(0,0) - A \cdot h - B \cdot k}{\sqrt{h^2+k^2}} \overset{?}{=} 0 , $$ where $A = \frac{\partial f}{\partial x}(0,0) = 1$ and $B = \frac{\partial f}{\partial y}(0,0) = 1$. You get $$ \lim_{(h,k)\to (0,0)} \frac{h^3+k^3 - (h+k)(h^2+k^2)}{(h^2+k^2)^{3/2}} , $$ which doesn't exist, since you get different values when approaching from different directions. So $f$ is not differentiable at the point $(0,0)$, despite having partial derivatives w.r.t. $x$ and $y$ there.

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