continuity and differentiability of function of two variables Let $f(x,y)$ be
$$f(x,y): \begin{cases} x & \text{for } y = 0\\ x-y^3\sin\left(\frac{1}{y}\right)& \text{for } y \neq 0\end{cases} $$
then check continuity and differentiability  at $(0,0)$.   

$$
f(x,y) = x - y^3\sin(.) = x + o(x,y)
$$
because $$\lim_{(x,y)\to (0,0)}\frac {y^3\sin(.)}{\sqrt{x^2 + y^2}} =0$$


*

*For continuity:
$$\lim_{(0,0)} f(x,y) = \lim_{(0,0)}x + \lim_{(0,0)}o(x,y) =0
$$
then $f$ is continunous in $(0,0)$.

*For differentiability:
Now identify with the definition of the derivative:
$$
f(x,y) = f(0,0) + Df(0,0)\cdot(x,y) + o(x,y)
$$
gives you the result:
$f$ is diffrentiable in in $(0,0)$ and $Df(0,0)\cdot(x,y) = x$.

