Determining when $f(x,y) = x^{4/3} \sin(y/x)$ ($x \ne 0$) is differentiable. Let $f$ be defined as follows:
$$
f(x,y) = \left\{
\begin{array}{lr}
x^{4/3} \sin(y/x) & \mathrm{if\ } x \ne 0 \\
0 & \mathrm{if\ } x = 0.
\end{array}
\right.
$$
I am asked to determine where $f$ is differentiable. Here is what I have so far:
When $x \ne 0$, $f$ is a composition of differentiable functions and so $f$ is differentiable. However, suppose I want to determine if $f$ is differentiable for $x = 0$ and any $y$; when I try to evaluate the partial derivative $f_x$ at $(0,y)$, I get
$$f_x = \lim_{h \to 0} \frac{f(h,y) - f(0,y)}{h} = \lim_{h \to 0} h^{1/3} \sin(y/h)$$
which I say is equal to $0$ because its magnitude is bounded above by $|h^{1/3}|$.
Similarly when I take the partial derivative with respect to $y$, I get $f(0,y+h) = f(0,y) = 0$. Then the partial derivatives exist and are equal, hence $f$ is differentiable everywhere.
Is this reasoning correct, or is there something I have overlooked?
 A: Existence of partial derivative does not guarantee the existence of derivative. But if each partial derivative is continuous at point $(a,b)$ then it has derivative at $(a,b)$. 
Unfortunately, partial derivative of $f$ respect to $x$ is not continuous, since
$$f_x(x,y)=\begin{cases}
\frac{4}{3}x^{1/3}\sin(y/x)-x^{-2/3}y\cos(y/x)&\text{if }x\neq 0\\
0&\text{if }x=0
\end{cases}$$
so $\lim_{h\to 0}f_x(h,1)$ is not even defined. 

In fact, it is not differentiable at $(x,y)=(0,0)$. If it is differentiable at $(x,y)=(0,0)$, then there is $a,b$ such that
$$\lim_{(h,k)\to (0,0)}\frac{h^{4/3}\sin(h/k)-ah-bk}{\sqrt{h^2+k^2}}=0$$
holds. But if we take $k=\sqrt{h}$ then
$$\lim_{h\to 0}\frac{h^{4/3}\sin\sqrt{h}-ah-b\sqrt{h}}{\sqrt{h^2+h}}=0$$
and you can check the limit in left side is not well-defined. 
Thanks to Jishu Das, I realize that my above argument is not correct. If we take $a=b=0$ then last equality in dashed line holds.
In fact, $f$ is differentiable everywhere. Its derivative is zero when $x=0$. It is given from following calculation:
$$\lim_{(x,y)\to(0,0)} \frac{|f(x,y)|}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)}\frac{|x^{1/3}\sin(y/x)|}{\sqrt{1+(y/x)^2}}\le \lim_{(x,y)\to(0,0)}|x^{1/3}| = 0$$
If $x\neq 0$, we can calculate its derivative directly, by evaluating componentwise partial derivative. Of course, we need to check that it is the derivative of $f$. It is given from the continuity of partial derivatives, as in the lecture note referred by Jishu.
A: f is differentiable everywhere in $R^2$. Refer to the third question of the pdf linked below.
https://www.math.ucdavis.edu/~hunter/m125b/m125b_midterm2_sample_solutions.pdf
