Show that the function $f(x,y)$ admits derivatives

Question

I have this function: $$f(x,y)=\frac{x^2y}{x^2+y^2}$$ when $(x,y)$ is not equal to $(0,0)$ and $$f(x,y)=0$$ when $(x,y)$ is equal to $(0,0)$. $$$$ What I have to do is:

1. Show that my function admits dervates after any direction in the point (0,0);
2. Find its partial derivates in (0,0);
3. Calculate it's derivate after the direction $s=(\frac{\sqrt3}{2},\frac{1}{2})$; $$$$ For 1. I have used this formulas to show that the function admits derivate after any direction in the point (0,0): $$\frac{df}{dx}(0,0)=\lim_{x\to 0} \frac{f(x,0)-f(0,0)}{x}$$ and for y $$\frac{df}{dy}(0,0)=\lim_{y\to 0} \frac{f(0,y)-f(0,0)}{y}$$ The first question comes here. How do I solve those limits? Can I just say that they are 0(I mean its $\frac{x^20}{x^2+0}$)? $$$$ For 2. I have just derivated the function, than i substituted $x,y$ with $0,0$. The second question comes here. I got this: $$\frac{df}{dx}(0,0)=\frac{2xy[1-x(x+y)]}{(x^2+y^2)^2}$$ and I don't think that I'm allowed to substitute $x,y$ with $0,0$. Can someone help me and put me back on the right track?(I also would realy appreciate a hint for 3.)

Answer 1

Let $s=(a,b)$ a direction.

The directional derivative in $(0,0)$ in this direction is given

by

$\lim_{t \to 0} \frac{f(t(a,b))-f(0,0)}{t}$.

For 1. , you have to show that the above limmit exists.

To 2.: for $\dfrac{\partial f}{\partial x}(0,0)$ let $(a,b)=(1,0)$ and for $\dfrac{\partial f}{\partial y}(0,0)$ let $(a,b)=(0,1)$