Squeeze Theorem and Second Partial Derivative of a Piecewise function

Question

$$f(x,y)=\arctan((xy^3)/(x^2+y^2))$$ for $(x,y) \ne(0,0)$ and $$0$$ for $(x,y)=(0,0)$. I have tried to prove its continuity at the point $(0,0)$ by using the Squeeze Theorem for $$y=x$$ and $$y=mx$$ such that $$\lim_{x\to 0}\arctan(x^2/2)\le\arctan((xy^3)/(x^2+y^2))\le\lim_{x\to 0}\arctan((x^2m^3)/(1+m^2))$$ so that the $$\arctan((xy^3)/(x^2+y^2))=0$$by the squeeze theorem. And calculate its second degree derivative at (0,0) $$f_{xy}(0,0)$$, I couldn't see how to calculate the second degree derivative since the function is a piecewise one, wouldnt it be just 0? Any help would be appreciated.

Answer 1

For each $x\in\Bbb R$, $|\arctan x|\leqslant|x|$ and therefore, if $(x,y)\ne(0,0)$,$$0\leqslant\left|\arctan\left(\frac{xy^3}{x^2+y^2}\right)\right|\leqslant\frac{|xy^3|}{x^2+y^2}=|xy|\frac{y^2}{x^2+y^2}\leqslant|xy|.$$So, since $\lim_{(x,y)\to(0,0)}|xy|=0$, it follows from the squeez theorem that$$\lim_{(x,y)\to(0,0)}\arctan\left(\frac{xy^3}{x^2+y^2}\right)=0.$$