I'm trying to study the following function : $$f(x,y) = \begin{cases}\dfrac{x^2y^2}{x^2+y^4} & (x,y)\neq (0,0)\\ 0 & (x,y)=(0,0) \end{cases}$$ I started by showing it's continuous at $(0,0)$ :
\begin{align*} \lim_{(x,y)\to (0,0)} f(x,y) &= \lim_{(x,y)\to (0,0)} \dfrac{x^2y^2}{x^2+y^4}\\ &= \lim_{r\to 0 } r^2 \dfrac{\cos^2 \theta \sin^2 \theta}{\cos^2 \theta + r^2 \sin^4\theta}\\ &= 0 \end{align*} It's $0$ because the $r^2 \sin^4\theta\to 0$ so as the whole expression, and I checked my answer using Wolfram alpha and it is true.
Well when it comes to check if the function is $\mathcal{C}^1$, I must prove that the partial derivatives exist and they're continuous : $$\partial_x f(0,0) = \lim_{x\to 0 } \dfrac{f(x,0)-f(0,0)}{x}=0$$ But : $$\lim_{(x,y)\to (0,0)} \partial_x f(x,y) = \lim_{(x,y)\to (0,0)} \dfrac{2xy^6}{(x^2+y^4)^2} $$ Its limit doesn't exist as Wolfram Alpha says, because : $$\dfrac{2r^5\cos\theta \sin^6\theta}{(\cos^2\theta+r^2\sin^4\theta)^2}$$ Isn't bounded, but according to my philosophy that I used to solve the continuity it is since $r^2\sin^4 \theta\to 0$ and the limit as well.
Any help to overcome this confusion ?
