Let $F:\mathbb{R}^2 \to \mathbb{R}^2$ given by $$F(x,y) = \begin{cases} \text{($x^2+x^2sin(1/x),y)$} &if\ x \neq 0\\ \text{$(0,y)$} & if\ x=0 \\ \end{cases}$$ Show that
i) $D_1F$ exists for every point in $\mathbb{R}^2$
ii)$D_2F$ exists and is continuous in some neighbourhood of $(0,0)$
iii) $F$ is differentiable in $(0,0)$
What I got: i) if $x \neq 0$ then we have $\dfrac{\partial f_1}{\partial x}(x,y) = 2x(1+sin(1/x))-cos(1/x)$ and $\dfrac{\partial f_2}{\partial x}(x,y)=0.$
if $x=0$ then $\dfrac{\partial f_1}{\partial x}(0,y) = 0$ and $\dfrac{\partial f_2}{\partial x}(0,y)=0$.
So $D_1F(x,y) = (2x(1+sin(1/x))-cos(1/x),0)$ if $x \neq 0$ and $D_1F(x,y) = (0,0)$ if $x=0$
ii) How can I show that $D_2F$ exists and is continuous in some neighbourhood of $(0,0)$?
iii) I have no clue about this one.
Any hint would be nice.
Edit: if $x \neq 0$ then $\dfrac{\partial f_1}{\partial y}(x,y) = 0$ and $\dfrac{\partial f_2}{\partial y}(x,y) = 1$
if $x=0$ then $\dfrac{\partial f_1}{\partial y}(0,y) = 0$ and $\dfrac{\partial f_2}{\partial y}(0,y) = 1$
Therefore $D_2F(x,y) = (0,1)\ \forall (x,y) \in \mathbb{R}^2$.Since it is a constant function, we have that is continuous at $(0,0)$(is this right?).
for iii) I know that $DF(0,0) = D_1F(0,0) + D_2F(0,0)$, but $$\dfrac{\partial f_1}{\partial x}(x,y) = \begin{cases} \text{$x^2+x^2sin(1/x)-cos(1/x)$} &if\ x \neq 0\\ \text{$0$} & if\ x=0 \\ \end{cases}$$ it's not a continuous function at $x=0$. Can I guarantee that F is differentiable in $(0,0)$ by only $D_2F(x,y)$ be continuous?