<s>[INST] Is $f:\mathbb{R}^2\to\mathbb{R}$ differentiable on $(0,0)$? Let $f:\mathbb{R}^2\to\mathbb{R}$ such that $f(x,y) = (x^2+y^2)\sin(\frac{1}{x^2+y^2})$ for $(x,y)\ne (0,0)$ and $f(x,y)=0$ for $(x,y)=(0,0)$. Is $f$ differentiable on $(0,0)$?
So let's first evaluate the partial derivatives:
$\frac{\partial f}{\partial x} = 2x\sin(\frac{1}{x^2+y^2}) + (x^2+y^2)\cos(\frac{1}{x^2+y^2})\frac{-2x}{(x^2+y^2)^2} = 2x\left( \sin(\frac{1}{x^2+y^2}) - \frac{\cos(\frac{1}{x^2+y^2})}{x^2+y^2}  \right)$
By symmetry; 
$\frac{\partial f}{\partial y} = 2y\sin(\frac{1}{x^2+y^2}) + (x^2+y^2)\cos(\frac{1}{x^2+y^2})\frac{-2x}{(x^2+y^2)^2} = 2y\left( \sin(\frac{1}{x^2+y^2}) - \frac{\cos(\frac{1}{x^2+y^2})}{x^2+y^2}  \right)$
 [/INST] By continuity $f(0,0)=0$ (because $0$ is the limit of $f(x,y)$ when $(x,y)\rightarrow (0,0)$ Now let us look at $$lim_{(x,y)\rightarrow (0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}$$ This limit is also $0$ and therefore $f$ is differentiable at the origin
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