Are the following function 1st derivatives continuous at (0,0)? Apologies: I've had to rephrase this question quite a few times
Suppose the following function:
$$f(x,y)=\frac{2x^3y^2}{x^4+y^2}$$
$$f(0,0)=0$$
Show it is differentiable/not differentiable at $(0,0)$
I attempted this question first by applying the definition of differentiability which yielded the following expression:
$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)}\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}$$
If the above equals 0, then the function is infact differentiable at $(0,0)$. I can't seem to find a counterexample which proves that it is discontinuous so I attempted to prove its continuity by applying the $\epsilon-\delta$ definition of limits. Unfortunately, no matter whether I express the function in terms of polarcoordinates or not, I cannot bound the following
$$|\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}| < \epsilon$$
for all $\epsilon$.
I'd be greatly appreciative for a solution to this kind of problem!

Credits to Kavi Rama Murthy for pointing out the $|2x^2y|<x^4+y^2$ inequality for me.
Note that:

*

*$(x^2-y)^2 \geq 0 \implies x^4 - 2x^2y + y^2 \geq 0 \implies x^4 + y^2 \geq 2x^2y$

*$(x-y)^2 \geq 0 \implies x^2 - 2xy + y^2 \geq 0 \implies x^2 + y^2 \geq 2xy \implies x^2+y^2 \geq xy \implies \sqrt{x^2+y^2} \geq \sqrt{xy}$
So it turns out the function is differentiable at (0,0). If $0 < \sqrt{x^2+y^2} < \delta$ then:
$$|\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}| = |\frac{(2x^2y)(\sqrt{xy})(\sqrt{xy})}{(x^4+y^2)\sqrt{x^2+y^2}}| = |\frac{2x^2y}{x^4+y^2}||\frac{\sqrt{xy}}{\sqrt{x^2+y^2}}||\sqrt{xy}| \leq (1)(1)\sqrt{xy} < \delta = \epsilon
$$
Thus the definition of limits is satisfied therfore $f(x,y)$ is differentiable at $(0,0)$
