Check differentiability of 2 variable function Let $f(x,y)=\frac{x^3y^3}{x^2+y^2}$ if $(x,y)\ne(0,0)$, and $f(x,y)=0$ if $(x,y)=(0,0)$.
I want to check if $f$ is differentiable at $(0,0)$. 
I know the function is continuous; will the difference of degree of numerator and denominator imply differentiability?
 A: Basically, yes.  You suspect, because of the degree, that the gradient vector is $\langle0,0\rangle$, so you use that in the definition.  For continuity, you checked that $\displaystyle\lim_{(x,y)\to(0,0)} f(x,y) = 0$; now you check that $\displaystyle\lim_{(x,y)\to(0,0)} \frac{f(x,y)}{\sqrt{x^2+y^2}} = 0$.  And because the degree of the numerator is so high, that's not hard.
(For the record, the general definitions that I'm using here are that $f$ is continuous at $(x_0,y_0)$ if $\displaystyle\lim_{(x,y)\to(x_0,y_0)} (f(x,y) - f(x_0,y_0)) = 0$, and that $f$ is differentiable at $(x_0,y_0)$ if $\displaystyle\lim_{(x,y)\to(x_0,y_0)} \frac{f(x,y) - f(x_0,y_0) - \mathbf{v}\cdot\langle{x-x_0,y-y_0}\rangle}{\lVert\langle{x-x_0,y-y_0}\rangle\rVert} = 0$ for some $\mathbf{v}$; here, $x_0$, $y_0$, $f(x_0,y_0)$, and $\mathbf{v}$ are all zero, so a lot of this simplifies.  If you're working with different definitions, then you may need to rewrite some of this, but it should amount to the same thing in the end.)
