Is $f(x,y)=\frac{y^3-\sin^3x}{x^2+y^2}$ differentiable at $(0,0)$? 
Define $f:\mathbb{R}^2\rightarrow \mathbb{R}$ by $f(x,y)=\displaystyle \frac{y^3-\sin^3x}{x^2+y^2}$ if $(x,y)\neq (0,0)$ and $f(0,0)=0$.
  My question is, is $f$ differentiable at $(0,0)$?

First I guessed it is not. So, I tried to prove that $f$ is not continuous at $(0,0)$ by showing the limits are different for two different paths approaching $(0,0)$. But could not find it easily. Then I tried proving it is differentiable. 
So I took $\displaystyle \lim_{(x,y)\rightarrow (0,0)}\frac{f(x,y)-f(0,0)-\nabla f(0,0).(x,y)}{\sqrt{x^2+y^2}}=\lim_{(x,y)\rightarrow (0,0)}\frac{x^3-\sin^3x}{(x^2+y^2)^{(3/2)}}$. But failed to prove that this limit is $0$. 
So, how could I determine whether $f$ is differentiable at $(0,0)$?
 A: Your approach is completely correct, but the execution is wrong. First of all, the gradient at the origin is not $(0,0)$; it should be $(-1,1)$. Look carefully at $f(x,0)$, and you'll see that 
$$\lim_{h\to 0} \frac{f(h,0)-f(0,0)}h = -1,$$
etc.  
So we need to decide if
$$\lim_{(x,y)\to (0,0)}\frac{\frac{y^3-\sin^3x}{x^2+y^2} - (-x+y)}{\sqrt{x^2+y^2}} = 0$$
or not. I'll let you work out the algebra, but you should need to determine whether 
$$\lim_{(x,y)\to (0,0)} \frac{xy(x-y)}{(x^2+y^2)^{3/2}} = 0;$$
what do you think?
A: So, because $\nabla f(0,0) = (-1,1)$you need to check if  $$\lim_{(x,y) \to (0,0)} \frac{x^3 - (\sin(x))^3 + xy(y-x)}{(x^2 + y^2)^\frac{3}{2}} = 0$$
We can take $(x,y) \to (0,0)$ along any path, so choose convergence along the positive $x-$axis, i.e fix $y=0$ and let $x \to 0+$
We get $\lim_{x\to 0+} -(\frac{sin(x)}{x})^3 = 0$
Now, set $y = kx$ and let $ x \to 0+$ The limit becomes 
$\lim_{y\to 0+} \frac{x^3 - \sin^3(x)+ x^3 (k^2 - k)}{x^3(1 + k^2)^\frac{3}{2}} \neq 0$ when (say) $k = 0.5$
So, the limit does not exist and the function is not differentiable. 
