$f(x,y)=\frac{x^3}{x^2+y^2}$ is not differentiable at $(0,0)$ 
Define $f(x,y)=\frac{x^3}{x^2+y^2}$ if $(x,y)\neq(0,0)$ and $f(x,y)=0$ for $(x,y)=(0,0)$
  Show that it is not differentiable at $(0,0)$

I figured out that both $f_x$ and $f_y$ exists and are discontinuous at $(0,0)$  but can't say anything about differentiability of $f(x,y)$ at $(0,0)$ and also $f(x,y)$ looks like a continuous function!

Were your function differentiable, it would be the case that if $f'(0;v)$ is the directional derivative at $0$ with direction $v$
$$f'(0;v+w)=f'(0;v)+f'(0,w)$$
Now let $v=(v_1,v_2)$ such that $v\neq 0$. Then
$$f'(0;v)=\lim\limits_{t\to 0}\frac{f(tv)}{t}=\lim_{t\to 0}\frac{t^3v_1^3}{t^3\lVert v\rVert ^2}=\frac{v_1^3}{\lVert v\rVert^2}$$
This is evidently not linear, hence your function cannot be differentiable at the origin.
