Partial derivative; is the function differentiable

[Throwback]

Homework Statement

[PLAIN]http://dl.dropbox.com/u/907375/Untitled.jpg

Homework Equations

Δz = f(a + Δx, b + Δy) - f(a, b)

[PLAIN]http://dl.dropbox.com/u/907375/Untitled2.jpg

The Attempt at a Solution

f_x(0,0)=lim(h->0)=0
f_y(0,0)=lim(h->0)=0
f(x,mx)=lim(h->0)=0
f(x,x)=lim(h->0)=0
f(x,0)=lim(h->0)=0
f(0,y)=lim(h->0)=0

This is apparently not enough. I have to also use the above equations to prove it's differentiable, but the only way I can think to express it in the above form is to convert to polar coordinates. That apparently requires some additional proof, which I'm unsure how to approach.

Converting to polar coordinates results in the following:

r^2*sin(1/r^2), but I'm unsure what else I have to do to account for switching to polar coordinates. Apparently it's easier to leave it is its original form (as a function of x and y) and use the above theorem, so here's that attempt:

((a+Δx)²+(a+Δy)²)sin(1/((a+Δx)²+(a+Δy)²))-(a²+b²)sin(1/(a²+b²))

((a+Δx)²sin(1/((a+Δx)²+(a+Δy)²))+(a+Δy)²sin(1/((a+Δx)²+(a+Δy)²)))-(a²sin(1/(a²+b²))+b²sin(1/(a²+b²)))

a²sin(1/((a+Δx)²+(a+Δy)²))+2a(Δx)sin(1/((a+Δx)²+(a+Δy)²))+(Δx)²sin(1/((a+Δx)²+(a+Δy)²))+a²sin(1/((a+Δx)²+(a+Δy)²))+2a(Δy)sin(1/((a+Δx)²+(a+Δy)²))+(Δy)²sin(1/((a+Δx)²+(a+Δy)²))-a²sin(1/(a²+b²)-b²sin(1/(a²+b²))

And now, thanks to that the sine part of the function, I'm lost. If it was simply sine of some constant, I'd be able to express ε1 and ε2, but because it's too a function of x and y, I'm pretty stuck.

 

[Nino MMP]

I don't even know if this is the correct approach, but I feel like it should be since the theorem requires a function of two variables and that's what I have.