A question about differentiability So, this is the last subject I have to study for my exam on Friday, and I still can not comprehend how to prove that a function is differentable.

$$f(x,y) = \begin{cases}
(x^2 + y^2) \cdot \sin \left(\frac{1}{\sqrt{x^2 + y^2}}\right) & \text{ if }(x,y) \neq (0,0),\\
0& \text{ if }(x,y) = (0,0).
\end{cases}$$
  Is $f(x,y)$ differentiable at $(x,y) = (0,0)$?

I know that I have to calculate the partial derivative according to the definition and then do another limit, but I do not know the formulas for these.
I am very, very sorry for asking such a general question, but I am really in trouble and don't seem to understand it.
I would appreciate very much a thorough answer.
 A: Hint Is $$f(x)=x^2\sin\frac{1}x$$ differentiable at the origin? You're now looking at $f\circ g(x,y)$ where $g(x,y)=\sqrt{x^2+y^2}$. What can you do now? Do you know how to apply the chain rule? Is $g$ differentiable (in the broader sense) at the origin? 
ADD Note that to know the partial derivatives of your function, one really needs to calculate the partial derivatives "by hand". That is $$\eqalign{
  & {\left. {\frac{{\partial f}}{{\partial x}}} \right|_{{x_0} = \left( {0,0} \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {h,0} \right) - f\left( {0,0} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} h\sin \frac{1}{{\left| h \right|}} = 0  \cr 
  & {\left. {\frac{{\partial f}}{{\partial y}}} \right|_{{x_0} = \left( {0,0} \right)}} = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0,h} \right) - f\left( {0,0} \right)}}{h} = \mathop {\lim }\limits_{h \to 0} h\sin \frac{1}{{\left| h \right|}} = 0 \cr} $$
So the partial deritvatives exist and are $0$. Thus, we now look at $$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left| {f\left( {x,y} \right) - f\left( {0,0} \right) - \nabla f\left( {0,0} \right) \cdot \left( {x,y} \right)} \right|}}{{\left\| {\left( {x,y} \right)} \right\|}}$$
where ${\left\| {\left( {x,y} \right)} \right\|}=\sqrt{x^2+y^2}$. The above is $$\mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \frac{{\left| {\left( {{x^2} + {y^2}} \right)\sin \left( {\frac{1}{{\sqrt {{x^2} + {y^2}} }}} \right)} \right|}}{{\sqrt {{x^2} + {y^2}} }} = \mathop {\lim }\limits_{\left( {x,y} \right) \to \left( {0,0} \right)} \left| {\sqrt {{x^2} + {y^2}} \sin \left( {\frac{1}{{\sqrt {{x^2} + {y^2}} }}} \right)} \right| = 0$$ so the function is differentiable. We used that $x\sin \frac 1 x\to 0$.
