I am trying to find out if the partial derivatives of my function $$f(x,y)=\begin{cases} 2\frac{x^3y}{x^2+y^2}-xy & \text{if } (x,y)\neq 0 \space \\ 0 & \text{if } (x,y)= 0 \space\\ \end{cases}$$ are continuous at all points. I have already calculated the derivatives for $\mathbb{R}\setminus\{0\}$ which are continous because they are combinations of polynomials. For $\frac{\partial}{\partial x}(0,0)$ and $\frac{\partial}{\partial y}(0,0)$ I have gotten $0$ as a result. In order to prove that my derivatives are continuous in their origin $(0,0)$, do I unterstand correctly that it is enough to show that $\lim_{(x,y) \to (0,0)} \frac{\partial}{\partial x} f(x,y) = \frac{\partial}{\partial x}(0,0) = 0$? And considering that is true, do I look at $\lim_{(x,y) \to (0,0)} \frac{\partial}{\partial x} f(x,y)$ for $x$ and $y$ seperately by setting one of them to a constant or can I just say $x = my$ for my prove? Any help is highly appreciated.
Best regards.
You need to show
$$\lim_{(x,y)\to (0,0)}\frac{\partial f}{\partial x}(x,y) = \frac{\partial f}{\partial x}(0,0)=0,$$
and the same for $\partial f/\partial y.$
For this it is not enough to consider limits on straight lines through $(0,0).$ There are examples where such limits along are all $0$ yet $f$ is not even continuous at $(0,0).$
Use your forumula for $\partial f/\partial x$ when $(x,y)\ne (0,0)$ to show that it has limit $0$ as $(x,y)\to 0.$ (Note that by definition, $(x,y)\to 0$ is the same as saying $(x^2+y^2)^{1/2}\to 0.$)