A Vector calculus - Prove a function is not differentiable at (0,0)

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The function f(x) is defined piecewise, with partial derivatives existing and continuity at (0,0). However, it is not differentiable at this point, as demonstrated by the lack of a limit in vector coordinates at (1,1). The initial attempt to use the definition of differentiability led to confusion, but ultimately, the vector coordinate approach clarified the situation. The conclusion is that despite the function's continuity and existing partial derivatives, it fails to meet the criteria for differentiability at (0,0). Understanding the vector coordinates method proved to be a straightforward solution.
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##f\left(x\right)=\begin{cases}\sqrt{\left|xy\right|}sin\left(\frac{1}{xy}\right)&xy\ne 0\\ 0&xy=0\end{cases}##
I showed it partial derivatives exist at ##(0,0)##, also it is continuous as ##(0,0)##
but now I have to show if it differentiable or not at ##(0,0)##.
According to answers it is not and they proved by showing the vector coordinates at ##(1,1)## does not have a limit.
But I dont want a proof like that, I tried using definition and got stuck...
I know my Linear transformation is basically 0. but still got in trouble of the definition
 
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Hi, is not needed anymore, at the end I managed to understand why they did vector coordinates, it is pratically trivial as I see it now
Thank thought :)
 

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