Showing that a function is differentiable

Question

So. I have to show that this functiun is differentiable outside of $(0,0)$ and also to calculate its differentiation. $$f(x,y)=xy\frac{x^2-y^2}{x^2+y^2}$$ when $(x,y)$ is not equal to $(0,0)$ and $$f(x,y)=0$$ when $(x,y)$ is equal to $(0,0)$. Also we have to show that the function is differentiable on $(0,0)$ too. $$$$ Firstly I wanted to see if the functiun is continuous e.g: $$\lim_{(x,y)\to (0,0)}{xy\frac{x^2-y^2}{x^2+y^2}}=0$$ so it is countinous.$$$$ Than I wanted to see if it is differentiable but I don't know how to continue, all I got is: $$\lim_{(x,y)\to (0,0)} \frac{f(x,y)-f(0,0)}{(x,y)-(0,0)}=\lim_{(x,y)\to (0,0)}\frac{xy\frac{x^2-y^2}{x^2+y^2}}{\sqrt{x^2+y^2}}$$ Can someone put me on the right track? And explain me what am I doing wrong? I don't know how to continue it..

Answer 1

Hint. By changing to polar coordinates one gets, as $(x,y) \to (0,0)$, $$ \left|\frac{f(x,y)-f(0,0)}{(x,y)-(0,0)}\right|=\left|\frac{xy\frac{x^2-y^2}{x^2+y^2}}{\sqrt{x^2+y^2}}\right|\le\rho\cdot|\cos \theta \cdot \sin \theta| \cdot|\cos^2 \theta-\sin^2 \theta|\le 2\rho \to 0 $$ yielding the desired differentiability.

Answer 2

Observe

$$\tag 1\left |\frac{xy\frac{x^2-y^2}{x^2+y^2}}{\sqrt{x^2+y^2}}\right| \le \frac{|xy|\cdot 1}{\sqrt{x^2+y^2}}.$$

Since $|xy| \le (x^2+y^2)/2,$ the limit in $(1)$ is $0.$