Given the function
$$f(x,y) = \begin{cases}y^2\arctan\left(\frac xy\right),& y \neq 0\\ 0, & y=0\end{cases}$$ I have to find $f_{yx}(x,y)$ and $f_{xy}(x,y)$. I notice that for $f_x(x,y)$ I obtain $0$ because it identically $0$in the whole "x axis" so $f_{xy}(x,y)$ is $0$.
Then for $f_y(x,y)$ I should see if it exists, then I can proceed to the second derivative. I obtain that $f_y(x,y)=0$ so $f_{yx}(x,y)=0$.
Is my solution correct or did I make some mistakes? Please if the solution is not correct, explain to me the right way to proceed.
Thank you in advance!
I shall interchange the rôles of $x$ and $y$. Due to symmetry we then can restrict to the halfplane $H$: $x\geq0$, and the function $f$ appears as $$f(x,y):=x^2{\rm Arg}(x,y)\quad\bigl((x,y)\in\dot H\bigr), \qquad f(0,0)=0\ .$$ For all points $(x,y)\in \dot H:=H\setminus\{(0,0)\}$ one has $|{\rm Arg}(x,y)|\leq{\pi\over2}$. Therefore $f$ is continuous on $H$. Note that $${\rm Arg}_x(x,y)={-y\over x^2+y^2}, \quad {\rm Arg}_y(x,y)={x\over x^2+y^2}\qquad\bigl((x,y)\in\dot H\bigr)\ .$$ Formal differentiation therefore gives $$f_x(x,y)=2x\,{\rm Arg}(x,y)-{x^2y\over x^2+y^2},\quad f_y(x,y)={x^3\over x^2+y^2} \qquad\bigl((x,y)\in\dot H\bigr)\ .$$ Furthermore one has $$f_x(0,0)=\lim_{x\to0+}{f(x,0)-f(0,0)\over x}=0,\qquad f_y(0,0)=\lim_{y\to0}{f(0,y)-f(0,0)\over y}=0\ .$$ Comparing with $(1)$ we can see that both $f_x$ and $f_y$ are continuous at $(0,0)$.
Formal differentiation produces then the second partials $$\left.\eqalign{f_{xx}(x,y)&=-{2xy(x^2+2y^2)\over(x^2+y^2)^2}+{\rm Arg}(x,y)\cr f_{xy}(x,y)=f_{yx}(x,y)&={x^2(x^2+3y^2)\over (x^2+y^2)^2}\cr f_{yy}(x,y)&=-{2x^3y\over(x^2+y^2)^2}\cr}\quad\right\}\qquad\bigl((x,y)\in \dot H\bigr).$$ None of them has a removable singularity at $(0,0)$.
Finally we have to check whether the second partials at $(0,0)$ exist. In turn we obtain $$f_{xx}(0,0)=\lim_{x\to 0+}{f_x(x,0)-f_x(0,0)\over x}=0,$$ then $$f_{xy}(0,0)=\lim_{y\to 0}{f_x(0,y)-f_x(0,0)\over y}=0\ ,$$ but $$f_{yx}(0,0)=\lim_{x\to 0+}{f_y(x,0)-f_y(0,0)\over x}=\lim_{x\to 0+}{x^2\over x^2}=1\ .$$ Finally $$f_{yy}(0,0)=\lim_{y\to 0}{f_y(0,y)-f_y(0,0)\over y}=0\ .$$