Showing a function has first and second partials everywhere and continuity of the partials Let $f \left( \begin{array}{ccc}
x \\
y \end{array} \right)=  \begin{cases} 
xy\dfrac{x^2-y^2}{x^2+y^2}  & \mbox{if $(x,y) \neq (0,0)$}\\
0 &    \mbox{if $(x,y) = (0,0)$} \end{cases}. $
I have this function and I want to show it's first and second partials exist everywhere, the first partials are continuous and the seconds are not. Computing the partials is one thing but what Do i do about continuity?
Here are the first partials:
away from $0$
$$D_1(f) = \large\frac{4x^2y^3 + x^4y-y^5}{(x^2+y^2)^2}$$
$$D_2(f) = \large\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$$
 A: I'll handle $D_1(f)$.
Let $(x,y)\neq (0,0)$.
Note that 
$$\begin{align} 
|D_1(f)(x,y)|&=\left|\dfrac {y\left(x^4+4x^2y^2-y^4\right)}{\left(x^2+y^2\right)^2}\right|\\
&\leq \dfrac {y\left(x^4+4x^2y^2+y^4\right)}{\left(x^2+y^2\right)^2}\\
&=\dfrac{y\left(\left(x^2+y^2\right)^2+2x^2y^2\right)}{\left(x^2+y^2\right)^2}\\
&=y+\dfrac{2x^2y^3}{\left(x^2+y^2\right)^2}.\\
\end{align}$$
Thus. to prove that $\lim \limits_{(x,y)\to (0,0)}\left(D_1(f)(x,y)\right)=0$, it is enough to prove that $$\lim \limits_{(x,y)\to (0,0)}\left(y+\dfrac{2x^2y^3}{\left(x^2+y^2\right)^2}\right)=0.$$
In turn, to prove this, it enough to prove that $$\lim \limits_{(x,y)\to (0,0)}\left(y\right)=0 \land \lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^2y^3}{\left(x^2+y^2\right)^2}\right)=0.$$
This is a consequence of
$$\dfrac{2x^2y^3}{\left(x^2+y^2\right)^2}=\dfrac{2x^2y^3}{x^4+2x^2y^2+y^4}\leq \dfrac{2x^2y^3}{2x^2y^2}=y\substack{(x,y)\to (0,0)\\\huge\longrightarrow} 0.$$
Therefore, if $D_1(f)(0,0)=0$, then $D_1(f)$ is continuous. If $D_1(f)(0,0)\neq 0$, it is discontinuous at the origin.
Since $D_1(f)(y,x)=-D_2(f)(x,y)$, the same conclusions follow immediately for $D_2(f)$.
