Is $ f(x,y) = \frac{|x|^{5/2} y}{x^4 + y^2} $ differentiable at $(0,0)$? 
Let $ f: \mathbb{R}^2 \rightarrow \mathbb{R} $ be a function defined as : 
$$
\begin{cases} 
f(x,y) = \frac{|x|^{5/2} y}{x^4 + y^2} , &  (x,y) \not= (0,0) \\ 
f(0,0) = 0 
\end{cases} 
$$ 
  
  
*
  
*Compute $ \frac{df}{dx}(0,0) $ and $\frac{df}{dy}(0,0)$.
  
*Is $f$ differentiable at $(0,0)$?
  


I have problem with question 2.
For $f$ to be differentiable, the partial derivatives must exist and be continuous.
They exist because we have: 
$ \frac{df}{dx}(0,0) = \lim_{x \to 0} \frac{f(x,0) - f(0,0)}{x} = 0$ 
$ \frac{df}{dy}(0,0) = \lim_{y \to 0} \frac{f(0,y) - f(0,0)}{y} = 0$
but I cannot compute the partial derivatives because of the absolute value.
How to know if the partial derivatives are continuous at $(0,0)$ ? 
 A: $f$ is differentiable at $(0,0)$ and the derivative is the always vanishing map as for $(x,y)\neq(0,0)$ you have
$$0 \le \left\vert \frac {f(x,y)}{\sqrt{x^2+y^2}} \right\vert = \frac{\vert x \vert}{\sqrt{x^2+y^2}} \frac{\vert xy \vert}{x^2+y^2}\sqrt{\vert x \vert}\le \frac{\sqrt{\vert x \vert}}{2} \to 0$$
As $(x,y) \to (0,0)$.
Take care! There is indeed a theorem stating that the derivative exists and is continuous if and only if the partial derivatives exist and are continuous. But a map maybe be differentiable even if the partial derivatives exist but are not continuous.
A: With your last edit the answer to 2) is negative. For differentiability we need by definition that
$$
\rho(x,y)=\frac{f(x,y)-f(0,0)-f'_x(0,0)x-f'_y(0,0)y}{\sqrt{x^2+y^2}}\to 0
$$
as $(x,y)\to(0,0)$, however, with $y=x^2$ we get for $x\to 0$
$$
\rho(x,x^2)=\frac{|x|^{1/2}x^4}{2x^4\cdot |x|\sqrt{1+x^2}}=\frac{1}{2|x|^{1/2}\sqrt{1+x^2}}\not\to 0.
$$
A: To calculate the partial derivatives, recall that the derivative of the absolute value is $|\cdot|' = \text{sign}$, on $\mathbb{R}\setminus \{0\}$.
Therefore for $(x,y) \ne (0,0)$ we have
$$\frac{\partial f}{\partial x}(x,y) = \frac{\frac{5}2(\operatorname{sign} x)|x|^{3/2}y(x^4+y^2) - |x|^{5/2}y\cdot 4x^3}{(x^4+y^2)^2}$$
$$\frac{\partial f}{\partial y}(x,y) = \frac{|x|^{5/2}(x^4+y^2) - |x|^{5/2}y\cdot 2y}{(x^4+y^2)^2}$$
E.g. for $x > 0$ and $y = x^2$ we get
$$\frac{\partial f}{\partial x}(x,x^2) = \frac{5|x|^{7.5} - 4|x|^{7.5}}{4x^8} = \frac1{4\sqrt{|x|}} \not\to 0$$
as $(x,y) \to (0,0)$ so $\frac{\partial f}{\partial x}$ isn't continuous at $(0,0)$.
This says nothing about differentiability, but the other answers show that $f$ is not differentiable at $(0,0)$.
