Consider the function
$$f(x, y) = \frac{x^4y^3}{x^8 + y^4}$$ not at the origin, but $0$ at the origin. I already provd the continuity of the function at the origin. I have to show if it's differentiable.
Considering some paths, I keep getting
$$\lim_{(x, y) \to (0, 0)} \frac{1}{\sqrt{x^2+y^2}} \frac{x^4y^3}{x^8 + y^4} =0$$
Yet when proving in serious way the differentiability, I get stuck. Here is what I did:
\begin{equation} \begin{split} \vert \frac{x^4y^3}{(x^8+y^4)\sqrt{x^2+y^2}} \vert & \leq \frac{x^4y^2|y|}{(x^8+y^4)\sqrt{x^2+y^2}}\\\\ & \leq \frac{x^4y^2|y|}{\sqrt{x^2+y^2}2x^4y^2} \\\\ & = \frac{|y|}{\sqrt{x^2+y^2}} \\\\ & \leq \frac{\sqrt{y^2}}{\sqrt{x^2+y^2}} \\\\ & \leq \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} \\\\ & = 1 \end{split} \end{equation}
- From line two to line three I used AM/GM: $x^8 + y^4 \geq 2\sqrt{x^8y^4} = 2x^4y^2$ and I reversed
So, am I wrong or is the function not differentiable at zero?
Let $r=\sqrt{x^2+y^2}.$ You already proved that $$\frac{|f(x,y)|}r\le\frac{|y|}r.$$
But when $y\ne0$ we also have $$\frac{|f(x,y)|}r\le\frac{r^4|y|^3}{r(0+y^4)}=\frac{r^3}{|y|}$$ hence $$\frac{|f(x,y)|}r\le\sqrt{\frac{|y|}r\frac{r^3}{|y|}}=r.$$ Alternatively (without using your upper bound): if $x,y\ne0,$
$$\frac{|f(x,y)|}r\le\frac{x^4|y|^3}{r(x^8+0)}=\frac{|y|^3}{rx^4} $$
$$\frac{|f(x,y)|}r\le\frac{x^4|y|^3}{r(0+y^4)}=\frac{x^4}{r|y|}$$ hence $$\frac{|f(x,y)|}r\le\sqrt[4] {\frac{|y|^3}{rx^4} \left(\frac{x^4}{r|y|}\right)^3}=\frac{x^2}r\le r.$$