Given $ E = (xy : xy \neq 0 ) $ . Let $ f : \mathbb R^{2}\longrightarrow \mathbb R$ be defined by $$f(x,y)=\cases{0,& if $xy=0$\\y\sin\left(\frac{1}{x}\right)+x\sin\left(\frac{1}{y}\right),& otherwise} .$$
Let $ S_1 $ be set of points in $ \mathbb R^{2}$ , where $f_x$ exists and $S_2 $ be set of points where $ f_y$ exists . Let $E_1$ be points where partial wrt x is continous and $E_2$ be set of points where $f_y$ is continous
I heed to determine what $S_i$ and $E_i$ are, $i=1,2 $
ATTEMPT : I calculated partial derivatives as
$f_x$ = $ -(y/x^{2}) \cos(1/x) + x\sin(1/y)$
$f_y = \sin1/x - (x/y^{2} )\cdot\cos (1/y) $
These both partials do not exist at X axis and Y axis and origin. At other points in xy plane they exist. Same for continuity. Is that enough, I'm not confident? Can anyone please elaborate? Thanks
let us look at $f_x.$ you have the formula $f_x = -\dfrac{y}{x^2}\cos(1/x) + \sin(1/y)$
it is clear that $f_x$ is continuous everywhere except possibly on the $x$ and $y$ axis. we will look at the axes in turn.
on the $x$-axis, $f$ is identically zero. so $f_x(x, 0) = 0$ for all $x.$
on the $y$-axis, we will use the definition of $f_x$ to compute $f_x.$
for $y \neq 0, f_x(0,y) = \lim_{h \to 0}\dfrac{f(h, y)}{h} =\dfrac{y \sin (1/h) + h\sin (1/y)}{h} = \sin(1/y)+ y\lim_{h \to 0}\dfrac{\sin (1/h)}{h}$ does not exist.
so the set $S_1 = \{(0,y) \colon y \neq 0 \}$
can you do the rest?