When is a function differentiable Let $f$ be a function 
$$
f(x,y)=\begin{cases}\quad0&(x,y)=(0,0)\\\dfrac{x^3y^2}{\left(x^2+y^2\right)^2} & (x,y)\neq(0,0).
\end{cases}$$
I know what $f$ is continous at the point because the limit of $f$ when $(x,y)\to(0,0)$ exist: did go to the origin by the $x$- and $y$-axis and all lines through origin. Since the limit exists, then $f$ is continuous and the partial derivatives exist in this point. But why isn't $f$ differentiable in $(0,0)$?
 A: As you have noticed, $f$ is continuous at $(0, 0)$, and partial derivatives exist:
$$
\left(\frac{\partial f}{\partial x}\right)_{0, 0}=\lim_{h\to 0}\frac{f\left(h, 0\right)-f\left(0, 0\right)}{h}=\frac{0-0}{h}=0
$$
Similarily, $D_y f(0, 0) = 0$. The definition of differentiability implies existence of the following limit:
$$
\lim_{(x, y)\to(0,0)}
\frac{f(x, y) - f(0, 0) - 
x \frac{\partial f}{\partial x}\left(0, 0\right) - 
y\frac{\partial f}{\partial y}\left(0, 0\right)}{\left|(x, y)\right|}=
\lim_{(x, y)\to(0,0)}\frac{x^3y^2}{\left(x^2+y^2\right)^{5/2}}
$$
I claim it does not exist. Let $x=r\cos \theta$, $y=r\sin \theta$ and let $r\to 0$. The above limit becomes
$$
\lim_{r\to 0}
\frac{r^5\cos\theta\sin\theta}{\left(r^2\cos^2\theta+r^2\sin^2\theta\right)^{5/2}}=
\lim_{r\to 0}\frac{r^5\cos\theta\sin\theta}{r^5}=\lim_{r\to 0}\cos\theta\sin\theta
$$
which clearly depends on $\theta$. 
Or, if you prefer, you can just explicitly choose two different directions and show they give different answers, e.g. $(x, y) = (0, t)$, $(x, y) = (t, t)$.
