prove that $f(x,y) = \frac{|x|\cdot y^2}{|x|+y^2}$ is differentiable at $(0,0)$

prove that the function defined by $f(x,y) = \frac{|x|\cdot y^2}{|x|+y^2}$ for $(x,y)\neq(0,0)$ and $f(0,0)=0$ is differentiable at $(0,0)$

I proved that the partial derivatives at $(0,0)$ exist (they both equal zero) but I have trouble proving that they are continous. How do I show that this limit $$\lim \limits_{(x,y)\to(0,0)} f_x(x,y) = \lim \limits_{(x,y)\to(0,0)} y^2\cdot \frac{2x+y^2sgn(x)}{\left ( |x|+y^2\right )^2}$$
is equal to $0$? and that's even before I looked into the other partial derivative, which I assume wouldn't be easy either.

• Choose two sequences, one approaching 0 from left and one from the right. If they coincide...
– wueb
Jul 14, 2017 at 18:39
• Note that $f_x=\dfrac{y^2\textbf{sgn}(x)}{(|x|+y^2)^2}$. Jul 14, 2017 at 18:55
• @JohnWaylandBales should be $y^4$ in the numerator, not $y^2$ Jul 14, 2017 at 18:58
• @eyeballfrog Correct, it was a typo but too late to correct. Thanks! Jul 14, 2017 at 19:01

after $AM-GM$ we have $$\frac{|x|+y^2}{2}\geq |y|\sqrt{|x|}$$ form here we get $$\frac{|x|y^2}{|x|+y^2}\le \frac{|x|y^2}{2|y|\sqrt{|x|}}=\frac{1}{2}|x|^{3/2}|y|->0$$
• OP: Please reread the definition of differentiability, then explain why your post does not even address the differentiability of $f$. // @upvoter Please explain your vote.
You know that the differential, if it exists, is the zero map. Now what you need is that $$\lim_{(x,y)\to(0,0)}\frac{1}{\sqrt{x^2+y^2}}\left|\frac{|x|\cdot y^2}{|x|+y^2}\right|=0$$ On the other hand, $$\frac{|y|}{\sqrt{x^2+y^2}}\le1 \qquad \frac{|x|}{|x|+y^2}\le1$$ so $$\frac{1}{\sqrt{x^2+y^2}}\left|\frac{|x|\cdot y^2}{|x|+y^2}\right| \le |y|$$