Proving the derivative of the multivariable function $f(x,y)=\frac{x^2y}{x^2+y^2}$ exists. 
Given is the function $f:\mathbb{R}^2\rightarrow\mathbb{R}$ with $f(0,0)=0$ and $f(x,y)=\dfrac{x^2y}{x^2+y^2}$ for $(x,y)\neq(0,0)$. 

I understand the function is continuous at $(0,0)$ and its partial derivatives exist at $(0,0)$. I need to prove the function is differentiable at $(0,0)$. The problem is that I'm unsure of what "derivative" means in this context. Is it the total derivative?
So far I thought of
$$\frac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}=\frac{\frac{x^2y}{x^2+y^2}}{\sqrt{x^2+y^ 2}}\leq\frac{x^2y}{(x^2+y^2)y}=\frac{x^2}{x^2+y^2}\leq\frac{x^2}{x^2}=1.$$Therefore $\lim_{(x,y)\rightarrow(0,0)}\dfrac{f(x,y)-f(0,0)}{\sqrt{x^2+y^2}}\leq1$. Is this enough to prove the derivative exists in $(0,0)$?
 A: The derivative is a linear map $L\colon\mathbb{R}^2\longrightarrow\mathbb R$ such that$$\lim_{(x,y)\to(0,0)}\frac{\bigl|f(x,y)-f(0,0)-L(x,y)\bigr|}{\|(x,y)\|}=0.$$Since it is a linear map from $\mathbb{R}^2$ to $\mathbb R$, it is a map of the form $L(x,y)=\alpha x+\beta y$ and it can be proved that $\alpha=\frac{\partial f}{\partial x}(0,0)=0$ and that $\beta=\frac{\partial f}{\partial y}(0,0)=0$. So, if your function was differentiable at $(0,0)$, its derivative would be the null function. But that doesn't work, because this would mean that$$\lim_{(x,y)\to(0,0)}\frac{\bigl|x^2y\bigr|}{\sqrt{x^2+y^2}^3}=0,$$which is false; if $y=x>0$,$$\frac{\bigl|x^2y\bigr|}{\sqrt{x^2+y^2}^3}=\frac{x^3}{2\sqrt2x^3}=\frac1{2\sqrt2}.$$
A: Recall that if partial derivatives exist and are continuos we are done, indeed by the "Differentiability theorem" if all the partial derivatives exist and are continuous in a neighborhood of the point then the function is differentiable at that point.
If partial derivatives are not continuos at the point we can't yet conclude anything about differentiability. We need to check directly differentiability by definition that:
$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}=0$$
In this case since
$$f_x=\frac{2xy^3}{(x^2+y^2)^2} \quad f_y=\frac{x^2(x^2-y^2)}{(x^2+y^2)^2}$$
are not continuos at $(0,0)$, by the definition we obtain
$$\lim_{(h,k)\rightarrow (0,0)} \frac{\| f(h,k)-f(0,0)-(f_x(0,0),f_y(0,0))\cdot (h,k)\|}{\| (h,k)\|}=\frac{|h^2k|}{\sqrt{(h^2+k^2)^3}}\neq 0$$
