A function that is continuous at the origin but not differentiable there Show that $f(x,y)=\dfrac{xy^2}{x^2+y^2}$ (with $(x,y)\not=(0,0)$ and $f(0,0)=0$) is continuous but not differentiable at $(0,0)$.
I tried to show continuity with an $\epsilon -\delta$ argument but I don't know how to factorize the expression so that I can have something useful. 
For differentiability I think I should show that the partials are not continuous at $(0,0)$. But finding the partials is also painful. 
 A: For continuity, a common trick is to express $f(x,y)=g(x,y)h(x,y)$ where $g$ has  limit $0$ at $(0,0)$ and $h$ is bounded in a punctured neighborhood of $(0,0)$. This is easy here:
$$
g(x,y)=x,\qquad h(x,y)=\frac{y^2}{x^2+y^2}
$$
because it's obvious that $0\le h(x,y)\le 1$ for all $(x,y)\ne(0,0)$.
Differentiability doesn't imply continuity of the partial derivatives; in some sense it's the other way round.
This function has partial derivatives at $(0,0)$ and both are zero:
$$
\lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}=
\lim_{h\to0}\frac{1}{h}\frac{h\cdot0^2}{h^2+0^2}
=0
$$
and
$$
\lim_{h\to0}\frac{f(0,0+h)-f(0,0)}{h}=
\lim_{h\to0}\frac{1}{h}\frac{0\cdot h^2}{h^2+0^2}
=0
$$
So, if the function is differentiable at $(0,0)$, its differential must be zero. Can you go on?
A: By the inequality
$$|f(x,y)|=\dfrac{|x|y^2}{x^2+y^2}\leq \dfrac{|x|(x^2+y^2)}{x^2+y^2}=|x|$$
we see that $f$ is continuous at $(0,0)$.
For the differentiability, your thinking is correct so do not hesitate and do not be afraid of calculus
A: We guess the limit is $0$ and so should verify that:
$$\forall\epsilon>0~\exists\delta>0,~\forall(x,y)\left(0<||(x,y)-(0,0)||<\delta\Longrightarrow\big|\frac{xy^2}{x^2+y^2}-0\big|<\epsilon\right)$$
When $0<||(x,y)-(0,0)||<\delta$ so $\sqrt{x^2+y^2}<\delta$ and from this we get $$|x|<\delta,~~|y|<\delta$$ Now if we set $z=\text{max}(|x|,|y|)$ then $z<\delta$ and also $$|\frac{xy^2}{x^2+y^2}-0|=\frac{|x||y|^2}{|x|^2+|y|^2}<\frac{z^3}{z^2+0}=z<\delta$$ So, it is enough to choose $\epsilon=\delta$. This shows that the function is continuous at the origin.
A: For proving continuity you may put $x = r\cos\theta$ and $y= r \sin\theta$. Change the function to polar coordinates. As $(x,y) \rightarrow (0,0)$, $r \rightarrow 0$. This may be a method for showing continuity.
For your second question related to differentiability see the following.
For any $v \in \mathbb{R}^2$, and $t \in \mathbb{R}$ $f(tv) = t f(v)$.
From $\epsilon , \delta$ definition you may easily show $D_{(1,0)} f(0,0) = D_{(0,1)} f(0,0) = 0$ where $D_{(1,0)}$ and $D_{(0,1)}$ are partial derivative along $x$ and $y$ axis, respectively.
But $D_{(1,1)}f(0,0) = \frac{1}{2}$. 
So the function is not differentiable at $(0,0)$.
