Differentiability with non continuous partials (origin) The function   $$f(x,y)= \frac{x^{2}y^{2}}{(x^{2}+y^{4})} \quad if \quad (x,y) \neq (0,0)$$
$$f(0,0)=0$$
In order to study it's differenciability at the origin, I've studied if the partial are continuous, and they are not. However, studying the limit $$\lim_{(x,y)\rightarrow(0,0)} \frac{\parallel f(x,y)-f(0,0)-\frac{\partial f}{\partial x}(0,0)x-\frac{\partial f}{\partial y}(0,0)y\parallel}{\parallel(x,y)\parallel }$$ 
I plugged in the limit of the expression of the partial and then took the limit out, because it's already the limit approaching the origin. Now, why should I've plugged 
$$\frac{\partial f}{\partial x}(0,0)=0 \quad and \quad \frac{\partial f}{\partial y}(0,0)=0$$  in the expression?  
I understand that $0$ is a linear function that makes that limit $0$, so it is the derivative at the origin. However, the partial are not continuous at the origin, how can I assume a value at the origin?
 A: To find the derivative of this function, you have make sure that the limit $$\lim_{(h_1,h_2)\rightarrow(0,0)} \frac{\parallel f(x_{0}+h_{1},y_{0}+h_{2})-f(x_0,y_0)-Ah\parallel}{\parallel(h_1,h_2)\parallel }\rightarrow0$$ where $A\in\mathbb{R^{1\times2}}$ and $$A=\left[\frac{\partial f}{\partial x}(x_0,y_0)\quad \frac{\partial f}{\partial y}(x_0,y_0)\right]=[0\;0]$$ and $$Ah=0$$ So, the function is Frechet differentiable if $$\lim_{(h_1,h_2)\rightarrow(0,0)} \frac{\parallel f(h_{1},h_{2})-f(0,0)\parallel}{\parallel(h_1,h_2)\parallel }\rightarrow0$$ 
A: If the function have continous partial derivate, that's a sufficient condition for the differentiability at the point, but is not necessary. So, there are functions with discontinous partial derivate at the origin, but they can be differenciability at the origin. If you want study the differentiability of a function, you can use the Taylor's formula
$f(\vec{a}+h\vec{y})-f(\vec{a})=\vec{\nabla}f(\vec{a})||\vec{y}||+||\vec{y}||E(\vec{a},\vec{y})$
Where $E(\vec{a},\vec{y}) \rightarrow 0$ when $||\vec{y}|| \rightarrow 0$ for any vector $y$ such that $h||\vec{y}|| \in B(\vec{a},r)$
That's the definition of a differentiable function at the point $\vec{a}$
