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For single-variable functions, we have that differentiability implies continuity. However, this is not the case with the following task, which examplifies this very well:

$$f(x,y) = \begin{cases}\frac{2xy}{x^2+y^2},\:\: (x,y) \ne (0,0)\\ 0,\:\: (x,y) = (0,0)\end{cases}$$ is not continuous at $(0,0)$, as the partial limits differ from the case $x = y$, where the limit is $1$. However, it is still differentiable here for both variables:

$f_x(0,0) = \lim_{h \to 0} \frac{0}{h^2} = 0$

By a similar argument, $f_y(0,0) = 0$.

We have established that a function may be partially differentiable although it is not continuous. While I do understand that this follows from the definition of partial derivatives, I am asking for an intuitive explanation based on properties of derivatives (slopes, normal lines etc.)

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    $\begingroup$ Note: $f$ is partially differentiable, it is not differentiable at $0$. Differentiability implies continuity also in higher dimensions, partial differentiability doesn't. $\endgroup$ Commented Mar 28, 2014 at 20:09
  • $\begingroup$ Remember that if the partial derivatives exist and are continuous, only then the function is differentiable. And differentiability does imply continuity. $\endgroup$ Commented Mar 28, 2014 at 20:11
  • $\begingroup$ @SandeepThilakan "Hence, the existence of partial derivatives does not imply that a function of several variables is continuous. This is in contrast to the single-variable case." Adams and Essex, Calculus: A Complete Course, 8th ed. $\endgroup$ Commented Mar 28, 2014 at 20:15
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    $\begingroup$ @AndrewThompson: Note the crucial word "continuous". $\endgroup$ Commented Mar 28, 2014 at 20:16
  • $\begingroup$ @SandeepThilakan: "Only then" should be just "then", otherwise the statement is false. $\endgroup$ Commented Mar 28, 2014 at 20:17

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Not much surprising. In the definition of $f_x(0,0)$ only values of $f(x,0)$ are considered. Similarly to define $f_y(0,0)$ only values of $f(0,y)$ are considered. This explains why the limit of $f(t,t)$ can be anything. In fact also this function has the property of being derivable in $(0,0)$ but not continuous: $$ f(x,y) = \begin{cases} 0 & \text{if $x=0$ or $y=0$}\\ 1 & \text{otherwise} \end{cases} $$

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I'll give you just a further example, which maybe could help you to clarify what's going on. In $\mathbb{R}^{2}$ consider the set $A=\lbrace \left(x,x^{2}\right)|x>0\rbrace$. Then define a function $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ as

$x \mapsto 1$ if $x \in A$,

$x \mapsto 0$ otherwise

Then this function is clearly not continuous in $\left(0,0\right)$, but it has any directional derivative in $\left(0,0\right)$, and they are all $0$. I make you this example because in general directional derivatives measure the behaviour of the function you cut along the line you choose. The point is that things can "approach badly" to the point you are considering along a "not straight" direction.

Note that you can modify this example to have $f$ continuous in $\left(0,0\right)$, with all directional derivatives, but not differentiable (for instance define $f$ on $A$ as $x$ and as $0$ out of $A$).

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I'll also give you just a further example, which maybe could help you to clarify what's going on.
Let $ f(x) := \begin{cases} 0 & \text{if $x$ is an irrational number,}\\ 1 & \text{if $x$ is a rational number.} \end{cases} $
Let $g(x,y):=f(x)\cdot f(y).$
If $a$ or $b$ is an irrational number, then $g(a,b)=0.$
If both of $a$ and $b$ are rational numbers, then $g(a,b)=1.$
Suppose that both of $a$ and $b$ are irrational numbers.
Then,
$\lim_{h\to 0}\frac{g(a+h,b)-g(a,b)}{h}=\lim_{h\to 0}\frac{0-0}{h}=\lim_{h\to 0}\frac{0}{h}=0.$
$\lim_{h\to 0}\frac{g(a,b+h)-g(a,b)}{h}=\lim_{h\to 0}\frac{0-0}{h}=\lim_{h\to 0}\frac{0}{h}=0.$
So, $g$ is partially differentiable at $(a,b)$.
But in any neighborhood of $(a,b)$, there are infinitely many points $(c,d)$ such that both $c$ and $d$ are rational numbers.
And in any neighborhood of $(a,b)$, there are infinitely many points $(c,d)$ such that $c$ or $d$ is an irrational number.
So, $g$ is not continuous at $(a,b)$.

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