Finding all the points (x,y) where $f(x,y)=\sqrt{|x^3y|}$ is differentiable. Finding all the points (x,y) where $f(x,y)=\sqrt{|x^3y|}$ is differentiable.
I started off with this:
If a function is differentiable then:
$f(x_0+\Delta x,y_0+\Delta y)=f(x_0,y_0)+\frac {\partial f(x_0,y_0)}{\partial x}\Delta x+\frac {\partial f(x_0,y_0)}{\partial y}\Delta y+\epsilon\sqrt{\Delta y^2+\Delta x^2}$
So what came to my head would be isolating $\epsilon$, And doing a limit where $\Delta x\to0$,$\Delta y \to 0$ and find when it equals to $0$. But the calculating it seems a bit problematic. Especially one that obsolete value.
Any tips?
 A: I suppose that "differential" means "differentiable". In this answer I would like to show the steps towards the solution, rather than the explicit computations. These can be added if requested.
The function $f$ defined on the open subset $\Omega$ in $\mathbb R^2$ is said to be  differentiable at $(x,y)\in \Omega$ if there exists a linear operator $A:\mathbb R^2\rightarrow \mathbb R$ s.t. for all $h=(h_1,h_2)\in\mathbb R^2$ with $(x,y)+(h_1,h_2)\in\Omega$
$f(x+h_1,y+h_2)-f(x,y)=A(h)+O(\|h\|^2)$.
If the function $f$ is differentiable at $(x,y)$, then $A(h)=\langle \nabla (f),h\rangle$, denoting by $\nabla(f)$ the gradient of $f$ at $(x,y)$. Moreover the function $f$ is continuous at $(x,y)$.
For these reasons we need to 


*

*Find those points (if any) where $f$ is not continuous; compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$, isolating the singularities of $\nabla(f)$ (if any).
To do so it is convenient to  consider the cases $f=\sqrt{x^3y}$ if $x^3y\geq 0$ and $f=
\sqrt{-x^3y}$ if $x^3y<0$. Draw the locus $x^3y=0$ in $\mathbb R^2$ to support your analysis.

*Check the limit $\lim_{\|h\|\rightarrow 0}\frac{f(x+h_1,y+h_2)-f(x,y)-\langle \nabla (f),h\rangle}{\|h\|}=0$, for all those $(x,y)$ for which $\nabla (f)$ exists. 
I hope these "guidelines" can help.
A: (i) When $x_0y_0\ne0$ then ${\rm sgn}(xy)$ is constant and $\ne0$ in some neighborhood $U$ of $(x_0,y_0)$. It follows that in computing $f_x$ and $f_y$ we may use the standard rules of calculus. When $x_0y_0>0$ we obtain
$$f_x(x,y)={3\over2}\sqrt{\mathstrut xy},\quad f_y(x,y)={1\over2}\sqrt{x^3\over y}\qquad\bigl((x,y)\in U\bigr)\ ,$$
and as these are continuous in $U$ the given function is continuously differentiable in $U$.
(ii) Consider a point $(0,y_0)$ on the $y$-axis.  We have
$$\eqalign{|f(\Delta x, y_0+\Delta y)-f(0,y_0)|&=\sqrt{|\Delta x|^3\>|y_0+\Delta y|}\cr&\leq |\Delta x|^{1/2}\sqrt{|\Delta x|^2(|y_0|+1)} \cr
&\leq |\Delta x|^{1/2}\sqrt{|y_0|+1}\sqrt{|\Delta x|^2+|\Delta y|^2}\ . \cr}$$
It follows that
$${|f(\Delta x, y_0+\Delta y)-f(0,y_0)|\over\sqrt{|\Delta x|^2+|\Delta y|^2}}$$
converges to $0$ when $(\Delta x,\Delta y)\to(0,0)$. Therefore $f$ is differentiable at $(0,y_0)$, and $df(0,y_0)=0$.
(iii) Consider a point $(x_0,0)\ne(0,0)$ on the $x$-axis. We have
$${f(x_0,y)-f(x_0,0)\over y}=|x_0|^{3/2}\ |y|^{-1/2}\ {\rm sgn}(y)\qquad(y\ne0)\ .$$
We conclude that $f_y(x_0,0)$ does not exist. A  fortiori $f$ is not differentiable at $(x_0,0)$.
To sum it all up: The function $f$ is differentiable at all points $(x,y)\in{\mathbb R}^2$, except at the points $(x_0,0)$ with $x_0\ne0$.
$\bigl[\ $The OP didn't ask for continuous differentiability. In this regard from
$$\lim_{(x,y)\to(0,y_0)} f_x(x,y)=\lim_{(x,y)\to(0,y_0)} f_y(x,y)=0\qquad(y_0\ne0)$$
it follows together with (ii) that $f$ is continuously differentiable also at all points $(0,y_0)$ with $y_0\ne0$.$\ \bigr]$
