Show that $f$ is not differentiable at $(0,0)$ - $\frac{x_1^2x_2}{x_1^2+x_2^2}$ 
Let the function $f: \mathbb{R^2} \to \mathbb{R}$ defined as 
$$f(x) =   
\begin{cases}   
\frac{x_1^2x_2}{x_1^2+x_2^2},     
 & \quad \text{if } (x_1,x_2) \not= 0  \\ 
0,  & \quad \text{if } (x_1,x_2) = 0
 \\    \end{cases} $$
Show that $f$ is not differentiable at $(0,0)$.

I think I can use the directionnal derivative or the polar coordinate, but I don't know how.
Is anyone is able to explain to me how I can use these properties? If I can't, could you propose to me an alternative?
P.S. Please, don't try to use a very specific analysis theory; I am only an undergraduate student (bachelor).

To check whether a function $f: \Bbb{R^2}\to \Bbb{R}$ is differentiable at $(0,0)$, we need to find a linear map $Df(x_0)$, called differential.
Now, let's take $u=(u_1,u_2)\in \Bbb{R^2-\{(0,0)\}}$. The directional derivatives along $u$ are:
$$D_uf(0,0)=\lim_{t\to 0}\frac{f(tu_1,tu_2)-f(0,0)}{t}=\lim_{t\to 0}\frac{f(tu_1,tu_2)}{t}=\lim_{t\to 0}\frac{t^3u_1^2u_2}{t^3(u^2_1+u^2_2)}=\frac{u^2_1u_2}{u_1^2+u_2^2}$$
So directional derivatives exist, and in particular, for $u=(1,0)$ and $u=(0,1)$, partial derivatives exist at $(0,0)$ and are equal to $0$. So our main candidate for the differential is $Df(0,0)(u_1,u_2)=0$. Then, in order to prove the differentiability of $f$ at $(0,0)$, we must prove that
$$\lim_{(x_1,x_2)\to (0,0)}\frac{x_1^2x_2}{(x_1^2+x_2^2)\sqrt{x_1^2+x_2^2}}=0$$
We'll see that this does not happen. Let $x_1=r\cos\theta$, $x_2=r\sin\theta$, Then
$$\lim_{(x_1,x_2)\to (0,0)}\frac{x_1^2x_2}{(x_1^2+x_2^2)\sqrt{x_1^2+x_2^2}}=\lim_{r\to 0}\frac{r^3\cos^2\theta \sin\theta}{r^3}=\cos^2\theta \sin\theta$$ 
So the limit depends on $\theta$, and therefore, it doesn't exist, so $f$ is not differentiable at $(0,0)$.
