Define $f(x,y) = 0$ if $(x,y) = (0,0)$ and $f(x,y) = \dfrac{x y^2}{x^2+y^2}$ if $(x,y) \neq (0,0)$.
a) Show that $f$ is continuous at the origin.
b) Prove $(D_uf)(x)$ exists at $(0,0)$ for any direction $\vec{u}$.
c) Show $f$ is not differentiable at $(0,0)$.
No ideas about any of them. Any help?
Recall the $f: R^2 \to R$ is differentiable at $(a,b)$ iff $f$ can be expressed in the form $f(x,y) = f(a,b) + f_x(a,b)(x-a)+f_y(a,b)(y-b) + E(x,y)$ with the error term $E(x,y)$ satisfies: $\displaystyle \lim_{(x,y) \to (a,b)} \dfrac{E(x,y)}{\sqrt{(x-a)^2+(y-b)^2}}=0$. Thus you either prove the partials does not exist or the limit of $E$ is not zero. Let's find out what really happens to $f$. We have: $f_x(0,0) = \displaystyle \lim_{h \to 0} \dfrac{f(h+0,0)-f(0,0)}{h}=\displaystyle \lim_{h \to 0} = \displaystyle \lim_{h \to 0} \dfrac{0-0}{h}= 0=\displaystyle \lim_{k \to 0} \dfrac{f(0,0+k)-f(0,0)}{k} = \displaystyle \lim_{k \to 0} \dfrac{0-0}{k} = 0$. Since $f(0,0) = 0$, we have $E(x,y) = f(x,y) = \dfrac{xy^2}{x^2+y^2}$. Thus: $\displaystyle \lim_{(x,y) \to (0,0)} \dfrac{E(x,y)}{\sqrt{x^2+y^2}} = \displaystyle \lim_{(x,y) \to (0,0)} \dfrac{xy^2}{(x^2+y^2)^{\frac{3}{2}}} = \text{ does not exist }$. All that you need to prove it does not exist is to choose $2$ paths for example: $(x,y) = (t,t)$, and $(x,y) = (2t,t)$
Hints:
(a) $\dfrac1{x^2 + y^2} \le \dfrac1{y^2}$ for $y\ne 0$.
(b) $f(hu_1, hu_2) = h \dfrac{u_1u_2^2}{u_1^2 + u_2^2}$ for $h\ne 0, u\ne 0$.
(c) Is $D_u f(0)$ linear in $u$?
In general, if $f:\mathbb R^{2}\to \mathbb R$ is differentiable at $\vec x=(x,y)$ then for all $\epsilon >0$, there is a $\delta_{\epsilon } >0$ such that
$$\tag 1 \vert f(\vec x+\vec h)-f(\vec x)\vert -\nabla f(\vec x)\cdot \vec h\vert <\epsilon \vec \vert h\vert $$ whenever $\vert \vec h\vert <\delta_{\epsilon } $.
Set $\vec x =0$. Then, we have $f(\vec 0)=0$, and $\nabla f(\vec 0)=0$ so $(1)$ becomes $$\tag 2 \left | \frac{f(\vec h)}{\vec h} \right |<\epsilon$$ which amounts to, upon setting $\vec h=(h,k)$, $$\tag 4\left | \dfrac{hk^2}{\left ( h^2+k^2 \right )^{3/2}} \right |<\epsilon$$
Now if we take $h=k$, $(4)$ becomes $\frac{k^{3}}{2\sqrt 2k^{3}}=\frac{1}{2\sqrt 2}$ so that as soon as $\epsilon < \frac{1}{2\sqrt 2}$, there is no $\delta $ that will satisfy $(2)$ and we have a contradiction.
