Prove that the function is differentiable but not continuously differentiable at a given point The function given is (from $\mathbb{R}^2$ to $\mathbb{R}$)
$$f(x,y) = \begin{cases} (x^2 + y^2) \sin( 1 / \sqrt{x^2 + y^2}) )  &\text{ when } (x,y) \neq (0,0), \\ 
0 &\text{ when } (x,y) = (0,0) \end{cases}
$$
Prove that the function is differentiable at $(0,0)$ , but not continuously differentiable at $(0,0)$
I have tried rewriting $f$ as a composition of two functions thinking that would make the proof easier, but it led me to a dead end. I assume directly using the definition of differentiability at a single point might solve the problem. However, I'm finding it hard to implement.
Thank you!
 A: Note that 
\begin{align}
f_x(0,0) & =  \displaystyle{\lim_{h\rightarrow 0} \frac{f(h,0)-f(0,0)}{h}} \\
& = \displaystyle{\lim_{h\rightarrow 0} \frac{h^2\sin(1/|h|)-0}{h}} \\
& = \displaystyle{\lim_{h\rightarrow 0} {h\sin(1/|h|)}} \\
& = 0
\end{align}
since $-1\leq\sin(1/|h|)\leq 1$ for all $h\neq 0$.
On the other hand, we can obtain 
$$f_x(x,y)=2x\sin(1/\sqrt{x^2+y^2})-\dfrac{x}{\sqrt{x^2+y^2}}\cos(1/\sqrt{x^2+y^2}),\;\; \text{for $(x,y)\neq (0,0)$.}$$ Let's see $f_x(x,y)$ is not continuous at $(0,0)$ by giving a sequency $\{(x_n,y_n)\}_{n\in\mathbb{N}}$ such that $(x_n,y_n)\rightarrow(0,0)$ but $f(x_n,y_n)\nrightarrow f_x(0,0)$.
Put $(x_n,y_n)=\left(\frac{1}{(2n+1)\pi},0\right)$ for $n\in \mathbb{N}$, it is easy to see $(x_n,y_n)\rightarrow(0,0)$, and we have
\begin{align}
f(x_n,y_n) & = 2x_n\sin(1/\sqrt{x_n^2})-\dfrac{x_n}{\sqrt{x_n^2}}\cos(1/\sqrt{x_n^2})\\
& = 2x_n\sin (1/x_n)-\cos(1/x_n) \\
& = \frac{1}{(2n+1)\pi}\sin((2n+1)\pi)-\cos((2n+1)\pi) \\
& = \frac{1}{(2n+1)\pi}(0)-(-1) \\
& = 1
\end{align}
So, $f(x_n,y_n)\rightarrow 1$ and $1\neq 0 =f_x(0,0)$ implies $f_x(x,y)$ is not continuous at $(0,0)$. In a similar way we can show $f_y(x,y)$ is not continuous at $(0,0)$
