calculus
For example in the following function:
$$f(x,y)=
\begin{cases}
\frac{x^3+y^3}{x^2+y^2},(x,y)\neq 0\\ 0,(x,y)=0\end{cases}$$
The partial derivatives in $(0,0)$
for $x$ and $y$ are $1$
therefore the derivative exists and is continuous at a point
but the function is not differentiable in $(0,0)$.
How is it possible?
There are no directional derivatives in nearly all directions. Consider, in particular, along the line $y=x$. $f(x,y)$ is a constant times the absolute value function.
When a function of two variables is differentiable, then there is a tangent plane to the surface $z=f(x,y)$, and there are directional derivatives in all directions. This one doesn't have directional derivatives except in two directions, and there's no tangent plane to the surface $z=f(x,y)$.
First of all:
$\frac{\partial f(x,y)}{\partial x} = \frac{2xy(x^4+y^2)-4x^3(x^2y)}{(x^4+y^2)^2}$
$\frac{\partial f(x,y)}{\partial y} = \frac{x^2(x^4+y^2)-2y(x^2y)}{(x^4+y^2)^2}$
and neither of the partial derivatives $\frac{\partial f(x,y)}{\partial x}$, $\frac{\partial f(x,y)}{\partial y}$ exist at $f(0,0)$
This is not right. The partial derivatives are particular directional derivatives (using the vectors $e_1=(1,0)$ and $e_2=(0,1)$), and you have to show that it exist (by definition below).
(As you wrote) Given a vector $u=(u_1,u_2)$ with $\|u\|=1$ we define the directional derivative of a function $f$ at a point $p=(p_1,p_2)$ by $$\dfrac{\partial f}{\partial u}(p) = \lim_{t\to 0} \dfrac{f(p+tu)-f(p)}{t} = \lim_{t \to 0} \dfrac{f(p_1+tu_1,p_2+tu_2) - f(p_1,p_2)}{t}$$ if the limit exist.
You have to show that this limit exist for every $u\in \Bbb{R}^2$ such that $\|u\|=1$.
Edit: To show that $f$ isn't differentiable at $(0,0)$, you can calculate the limit through different paths and conclude that the limits hasn't the same value, like Svetoslav said in comments.
Best Answer
No, the partial derivatives are not continuous in $(0,0)$. For $(x,y) \ne (0,0)$, you have: $$f_x(x,y)=\dfrac{x\left(x^3+3y^2x-2y^3\right)}{\left(x^2+y^2\right)^2}$$ $$f_y(x,y)=\dfrac{y\left(y^3+3x^2y-2x^3\right)}{\left(y^2+x^2\right)^2}$$ And, as you correctly noticed, $f_x(0,0)=f_y(0,0)=1$. But: $$\nexists\lim_{(x,y)\to (0,0)}f_x(x,y)$$ $$\nexists\lim_{(x,y)\to (0,0)}f_y(x,y)$$ For instance, consider $f_x(0,y)$ and $f_x(x,0)$ and similar restrictions for $f_y$.