Prove $f(x,y)$ is not differentiable in ($0,0)$

Question

Prove $f(x,y)$ is not differentiable in ($0,0)$

$$f(x,y)= \begin{cases}\dfrac{|x|y}{\sqrt{x^2+y^2}}& \text{if } (x,y)\not =0\\ \\ 0&\text{if } (x,y)=0 \end{cases} $$

I try prove this by existence of the limit.

Let $y=x$ and $x\not = 0$, then

$$f(x,y)=\frac{|x|x}{\sqrt{x^2+x^2}}=\frac{|x|x}{x\sqrt{2}}=\frac{|x|}{\sqrt{2}}$$

And $\lim_{x\rightarrow 0}f(x,y)=0.$

Moreover, let $x=my$ then $$f(x,y)=\frac{|my|y}{\sqrt{m^2y^2+y^2}}=\frac{m|y|y}{\sqrt{y^2(m^2+1)}}=\frac{m|y|}{\sqrt{(m^2+1)}}$$

This implies $\lim_{y\rightarrow 0}f(x,y)=0.$

I try other trayectories but don't work. Can someone help me with this?

Answer 1

If it were differentiable at $(0,0)$, the corresponding Jocobian is the partial derivatives at $(0,0)$, which you can compute the Jacobian is actually the zero map as well.

So you may try to get a contradiction with the existence of \begin{align*} \lim_{(x,y)\rightarrow(0,0)}\dfrac{1}{\sqrt{x^{2}+y^{2}}}\dfrac{|x|y}{\sqrt{x^{2}+y^{2}}}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{|x|y}{x^{2}+y^{2}}. \end{align*} For $y=mx$, $x>0$, then \begin{align*} \lim_{x\rightarrow 0^{+}}\dfrac{mx^{2}}{x^{2}+m^{2}x^{2}}=\lim_{x\rightarrow 0^{+}}\dfrac{m}{1+m^{2}}=\dfrac{m}{1+m^{2}}, \end{align*} which varies along with $m$.

Answer 2

Check from the definition that the partial derivatives of $f$ at $(0,0)$ are $0.$ Thus if $Df(0,0)$ exists, it equals the $0$ linear transformation. It would then follow that all directional derivatives at $(0,0)$ are $0.$ But

$$\frac{f(x,x) -f(0,0)}{x} = \frac{|x|x/(\sqrt 2 |x|)}{x} = \frac{1}{\sqrt 2},$$

which does not $\to 0,$ contradiction. Hence $Df(0,0)$ does not xist.