Q:

How can I find these partial derivatives?

I'm reading a book which gives this function $f(x,y)=x^2y/(x^2+y^2)$ if $(x,y)\neq (0,0)$ and $f(0,0)=0$ as a $C^1$ function in $\mathbb R^2-\{(0,0)\}$, continuous in $(0,0)$ and it has the partial derivatives $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$. In fact the author wants to point out a function which has the partial derivatives in $(0,0)$, but it's not differentiable in $(0,0)$.
The problem is I can't find the partial derivatives $\frac{\partial f}{\partial x}(0,0)$ and $\frac{\partial f}{\partial y}(0,0)$ I think I should use the limits.
See:
$$\frac{\partial f}{\partial x}(x,y)=\frac{2xy(x^2+y^2)-2x^3y}{(x^2+y^2)^2}$$
and
$$\frac{\partial f}{\partial y}(x,y)=\frac{x^2(x^2+y^2)-2x^2y^2}{(x^2+y^2)^2}$$
They aren't defined at $(0,0)$. How can I proceed to find these partial derivatives at $(0,0)$?
Thanks

A:

The definition of the partial derivative at $(x,y)$ with respect to $x$ is
$$\frac{\partial f(x,y)}{\partial x}=\lim_{h\to 0}\frac{f(x+h,y)-f(x,y)}{h}$$
The analogous limit applicable to the partial derivative with respect to $y$ at $(x,y)$ is
$$\frac{\partial f(x,y)}{\partial y}=\lim_{k\to 0}\frac{f(x,y+k)-f(x,y)}{k}$$
Here, $x=0$ and $y=0$ while $f(0,0)=0$.  In addition, $f(h,0)=0$ and $f(0,k)=0$.  Thus
$$\left . \frac{\partial f}{\partial x}\right|_{0,0}=\lim_{h\to 0}\frac{0-0}{h}=0$$
and
$$\left . \frac{\partial f}{\partial y}\right|_{0,0}=\lim_{k\to 0}\frac{0-0}{k}=0$$
So, both partial derivatives are $0$ at the origin.
It is interesting to remark that while the partial derivatives exist at the origin, the partial derivatives are not continuous there.

