Question about partial derivative at a point. I'm currently studying Partial Derivatives and ran into this problem:
$f(x,y) = \begin{cases} \frac{x^2}{x^2+y^4}, & (x,y) \neq (0,0) \\ 0, & (x,y) = (0,0) \end{cases}$
In this question, we are supposed to find $\frac{\partial f}{\partial x}$ at $(0,0)$.
(Sorry if this question seems very easy, but I'm new to the material).
From what I know, we use the definition as follows:
$\frac{\partial f}{\partial x}(x_0,y_0) = \lim_{\delta x \to 0} \frac{f(x_0+\delta x,y_0) - f(x_0,y_0)}{\delta x}$.
At $(0,0)$ we have $f(x,y) = 0.$ Therefore,
$\lim_{\delta x \to 0} \frac{0 - 0}{\delta x} = 0$.
Is this correct? Or do we do the following:
$\lim_{\delta x \to 0} \frac{\frac{\delta x^2}{\delta x^2 + 0^4}-\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^4}}{\delta x}$
We can observe that $\lim_{(x,y) \to (0,0)} \frac{x^2}{x^2 + y^4}$ does not exist as it converges to different values at different paths (try $x = y$ and $x = y^2$). So, $\frac{\partial f}{\partial x}(0,0)$ does not exist.
Now I believe the first solution is correct because the question says that at $(0,0)$, the function is equal to zero and so we only consider the zero while finding the limit at that point.
But I'm not 100% sure. I talked about this with other students and I found conflicting answers.
Would it be possible for someone to tell me which answer is the correct one? A brief explanation would be very much appreciated.
Thank you.

By definition,$$\frac\partial{\partial x}f(x_0,y_0)=\lim_{h\to0}\frac{f(x_0+h,y_0)-f(x_0,y_0)}h.$$In particular$$\frac\partial{\partial x}f(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}h.$$But $f(0,0)=0$ and, for each $h\in\Bbb R\setminus\{0\}$, $f(h,0)=\frac{h^2}{h^2}=1$. Therefore, the limit$$\lim_{h\to0}\frac{f(h,0)-f(0,0)}h\quad\text{is}\quad\lim_{h\to0}\frac1h,$$which does not exist (in $\Bbb R$).
