Multivariable differentiability of piecewise function at a point

The question is as follows:

Let $F:\mathbb{R}^2\rightarrow\mathbb{R}$ be a function given by $$F(x,y)=\bigg\{\begin{array}{cc} x^3\sin(1/x)+y^2 & x\neq 0\\ y^2&x=0. \end{array}$$ Show that $F$ is differentiable at the point $(0,0)$.

The piecewise nature of this function is throwing me off. I know that if all the partial derivatives are continuous in a neighborhood of the point, then the function is differentiable there. But $\frac{\partial}{\partial x}(x^3\sin(1/x)+y^2)=x(3x\sin(1/x)-\cos(1/x))$ is not continuous near $(0,0)$. So my inclination is to use the limit definition of derivative. However, $\lim_{(h,k)\rightarrow(0,0)}\frac{h^3\sin(1/h)+k^2+h-k}{\sqrt{h^2+k^2}}$ does not exist. Am I able to completely ignore the first piece since the function is not defined by it at $(0,0)$?

You have $f_x(0,0)=f_y(0,0)=0$. Therefore, if $f$ is differentiable at $(0,0)$, then $f'(0,0)$ can only be the null function. On the other hand, asserting that $f'(0,0)$ is the null function is the same thing as asserting that$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.\tag1$$And this is indeed true, because if $\bigl\|(x,y)\bigr\|<1$ you have$$\bigl|f(x,y)\bigr|\leqslant|x|^3+y^2\leqslant x^2+y^2=\sqrt{x^2+y^2}^2$$and therefore$$\left|\frac{f(x,y)}{\sqrt{x^2+y^2}}\right|\leqslant\sqrt{x^2+y^2}.$$So, $(1)$ holds and $f'(0,0)$ is indeed the null function.
• @Atsina You are right: I didn't express myself correctly. What I meant to say was that $f'(0,0)$ could only be the null function and then I proved that this is indeed the case. I've edited my answer. I hope that everything is clear now. – José Carlos Santos Aug 15 '18 at 17:46
• Could you briefly elaborate on why $f_x(0,0)=f_y(0,0)=0$ gives that $f'(0,0)=$ null function $\Rightarrow f$ is differentiable at $(0,0)$? – Atsina Aug 15 '18 at 17:59
• @Atsina Since we are deling with the limit at $(0,0)$, only vectors with a small norm matter. And I need to have $|x|<1$, so that the inequality $|x|^3\leqslant x^2$ holds. – José Carlos Santos Aug 15 '18 at 18:14