I want to prove that $$ f(x,y)= \begin{cases} \frac{xy^2}{x^2+y^2} &\text{ if }(x,y)\neq (0,0)\\ 0 &\text{ if }(x,y)=(0,0) \end{cases} $$ is not differentiable at $(0,0)$.
I thought that I can prove that it is not continuous around $(0,0)$ but it certainly is!
So how can I prove that it is not differentiable?
On
At the origin the directional derivative in the direction of $(1,1)$ is $\frac 1 2$ whereas the derivative in the direction of $x-$ axis and $y-$axis are $0$. This implies that the derivative does not exist.
On
Another approach: assume that $f$ is differentiable at $(0,0)$. For each $v=(u,w) \in \mathbb R^2$ with $u^2+w^2=1$ we then have
$$(*) \quad \frac{\partial f}{\partial v}(0,0)=v \cdot gradf(0,0).$$
Since
$$ (**) \quad\frac{\partial f}{\partial v}(0,0)= \lim_{t \to 0}\frac{f(tv)-f(0,0)}{t}=\frac{uw^2}{u^2+w^2}=uw^2,$$
we see that $gradf(0,0)=(0,0)$ and therefor, by $(*)$
$$\frac{\partial f}{\partial v}(0,0)=0.$$
But if $u=w= \frac{1}{\sqrt{2}}$, then by $(**)$
$$\frac{\partial f}{\partial v}(0,0) \ne 0.$$
A contradiction !
Remark: the solution of Jose Carlos Santos shorter. But my solution shows that a function $f$ can have directional derivatives at $x_0$ in all directions $v$, without being differentiable at $x_0$.
On
Look at the directional derivative:
$$d_{\mathbf{v}}f=\lim\limits_{t\to 0}\frac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}$$
in your case you have for $\mathbf{x}=(0,0)$
$$d_{(a,b)}f=\lim\limits_{t\to 0}\frac{\frac{t^3ab^2}{t^2(a^2+b^2)}}{t}=\frac{ab^2}{a^2+b^2}$$ Thus the directional derivative exists. However, it is not a linear function, and therefore $f$ is not differentiable.
Since $f_x(0,0)=f_y(0,0)=0$, if $f$ was differentiable at $(0,0)$, $f'(0,0)$ would be the null function. Therefore$$\lim_{(x,y)\to(0,0)}\frac{\bigl|f(x,y)-f(0,0)\bigr|}{\sqrt{x^2+y^2}}=0,$$which means that$$\lim_{(x,y)\to(0,0)}\frac{|xy^2|}{(x^2+y^2)^{\frac32}}=0.$$However, this is false. See what happens when $x=y$.