Showing that $f$ is twice differentiable Let $f : \mathbb R^2 → \mathbb R$ defined by $f(x, y) := \frac{(x−1)^3 y^3} {(x−1)^2+y^2}$ if $(x, y) \not= (1, 0)$ and $f(1, 0) = 0$. 
Answer the following questions:


*

*We take for granted that f is twice differentiable at $(1, 1)$. Compute $D^2_{(1,1)} f(1, 2) · (1, −2)$.

*Show that $f$ is twice differentiable at $(1, 0)$.
My attempts:


*

*I Calculated it and I found that it is equal to $0$.

*Here, I think I have to show that lim$_{h→0}$ $\frac{||D_h f - D_{(1,0)}f - L.h||}{||h||}$ = $0$, where $L$ is a linear map, $L: \mathbb R^2 → \mathbb R$. But I don't know how to continue from there. Any help please? Thank you
 A: $f(1+h,k)-f(1,0)=\frac{h^3 k^3} {h^2+k^2}$ so we look for a linear transformation $Df(1,0):\mathbb R^2\to \mathbb R$ such that $\frac{\left |Df(1,0)(h,k)-\frac{h^3 k^3} {h^2+k^2}\right|}{\sqrt{h^2+k^2}}\to 0$ as $(h,k)\to 0.$ If we try the easiest one, namely, $Df(1,0)(h,k)=0$ we see that it works.
For the second derivative at $(1,0),$ first look at the general case: we have the following data:
$f:\mathbb R^2\to \mathbb R;\ x\mapsto f(x);\ Df:\mathbb R^2\to L(\mathbb R^2,\mathbb R);\ x\mapsto  Df(x)$ and $Df(x)$ is the linear transformation defined as you have done. 
Now then, $D^2f:\mathbb R^2\to L(\mathbb R^2,L(\mathbb R^2,\mathbb R))$ defined as follows: 
If $Df$ is differentiable at $x_0\in \mathbb R$ then there must exist a map $D^2f$ that sends $x_0\in \mathbb R^2$ to a linear transformation $D^2f(x_0),$ which in turn  satisfies
$Df(x_0+h)-Df(x_0)=D^2f(x_0)(h)+r(h)$ where $r(h)/\|h\|\to 0$ as $h\to 0.$
So, basically we want to calculate $Df(x_0+h)-Df(x_0)-D^2f(x_0)(h)$ and show that it is small whenever $h$ is. 
This is the same definition of derivative except now $\mathbb R$ is replaced by $L(\mathbb R^2,\mathbb R)).$ Notice, these maps are all elements of $ L(\mathbb R^2,\mathbb R)$ so to make sense of them, we have to evaluate them at an arbitrary $v\in \mathbb R^2:$
$Df(x_0+h)(v)-Df(x_0)(v)-D^2f(x_0)(h)(v)$
Now, it's easier to express the derivatives as $1\times 2$ matrices: we have $x_0=(1,0)$ so writing $h:=(h,k),$ 
$Df((1,0)+(h,k))=\begin{pmatrix}
f_x(1+h,k) & f_y(1+h,k))
\end{pmatrix}=\begin{pmatrix}
\frac{k^3\left(3\left(h\right)^2\left(\left(h\right)^2+k^2\right)-2\left(h\right)^4\right)}{\left(\left(h\right)^2+k^2\right)^2} & \frac{\left(3(1+h)^2k^2-6(1+h)k^2+3k^2+k^4\right)\left(h\right)^3}{\left(\left(h\right)^2+k^2\right)^2}
\end{pmatrix}$ 
and $Df((1,0))=0$ as we showed above. 
So, as in the first part, we look for a linear transformation $D^2f(1,0)$ such that 
$\frac{\|Df(1+h,k)-D^2f(1,0)(h,k)\|}{\sqrt{h^2+k^2}}\to 0$ as $(h,k)\to 0.$
If we try $D^2f(1,0)=0$ again we have 
$\frac{Df(1+h,k)(v_1,v_2)}{\sqrt{h^2+k^2}}=\frac{k^3\left(3\left(h\right)^2\left(\left(h\right)^2+k^2\right)-2\left(h\right)^4\right)}{\left(\left(h\right)^2+k^2\right)^{5/2}} v_1+ \frac{\left(3(1+h)^2k^2-6(1+h)k^2+3k^2+k^4\right)\left(h\right)^3}{\left(\left(h\right)^2+k^2\right)^{5/2}}v_2.$ 
Suping this over $\|v\|\le 1$ and letting $(h,k)\to 0$ shows that our guess was correct.
