Let $f(x,y)= \left\{ \begin{array} ((x+y)^2\sin(\frac{\pi}{x+y}) & if & x +y\not = 0 \\ \\ 0 & if & x+y=0 \\ \end{array} \right.$
Prove $f$ is differentiable in $(0,0)$.
My work:
Let $\overline{h}=(h_1,h_2)$
Suppose $d_{(0,0)}f(\overline{h})=\pi h_1+\pi h_2$ then
$$\lim_{\overline{h} \rightarrow \overline{0}}\left|\left|\frac{f((0,0)+(h_1,h_2))-f(0,0)-df(\overline{h})}{h}\right|\right|=\lim_{\overline{h} \rightarrow \overline{0}}\left|\left|\frac{f((h_1,h_2))-(\pi h_1+\pi h_2)}{h}\right|\right|=\lim_{\overline{h} \rightarrow \overline{0}}\left|\left|\frac{(h_1+h_2)^2\sin(\frac{\pi}{h_1+h_2})-(\pi h_1+\pi h_2)}{h}\right|\right|$$
Here I'm stuck.
Note: I suppose $df(\overline{h})=\pi h_1+\pi h_2$ because $df(\overline{h})=<\nabla f(0,0),(h_1,h_2)>$
If $\bigl\|(x,y)\bigr\|=r$, then $x=r\cos\theta$ and $y=r\sin\theta$, for some $\theta\in\mathbb R$ and then$$(x+y)^2=r^2\bigl(1+\sin(2\theta)\bigr).$$Therefore\begin{align}\left|\frac{(x+y)^2\sin\left(\frac\pi{x+y}\right)}{\sqrt{x^2+y^2}}\right|&\leqslant\frac{(x+y)^2}r\\&\leqslant\frac{r^2\bigl(1+\sin(2\theta)\bigr)}r\\&=r\bigl(1+\sin(2\theta)\bigr)\\&\leqslant2r.\end{align}So, if $L\colon\mathbb{R}^2\longrightarrow\mathbb R$ is the null function$$\lim_{(x,y)\to(0,0)}\left|\frac{f(x,y)-f(0,0)-L(x,y)}{\sqrt{x^2+y^2}}\right|=0.$$In other words, $D_{(0,0)}f$ is the null function.