Proving a function of 2 variables is diffable

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Homework Statement



i)Prove that f: [tex]R^2\rightarrow R[/tex]

f(x,y)= [tex]\frac{x^4 + y^4}{(x^2+y^2)^\alpha}[/tex] (x,y) =/= (0,0)
f(x,y)= 0 (x,y)=(0,0)

is differentiable on [tex]R^2[/tex] for [tex]\alpha[/tex]<3/2



Homework Equations



let f be a vector function from n variable sto m variables
i) f is said to be diffable at a point a (element) R^n iff there is an open set V containing a such that f: [tex]R\rightarrow R^m[/tex] and there is a T (element) L(R^n;R^m) such that the function
[tex]\epsilon[/tex](h) := f(a+h)-f(a)-T(h) satisfies [tex]\epsilon[/tex](h)/||h|| ---> 0 as h--->0

ii) f is said to be differentiable on a set E iff E is not empty and f is diffable at every point in E


The Attempt at a Solution



I actually have no clue how to do this problem. I had my wisdom teeth pulled the day he covered this in lecture. please HALP MEH!
 
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i still need help on this problem
 
Well, suppose that [itex]\alpha<3/2[/itex], and consider the partials of f(x,y) and see if they are continuous. If they are, then f is differentiable.

[Note: I haven't tried this.]
 
I don't know if this method would work.

fx = [tex]\frac{(x^2 + y^2)4x^3 - 2ax(x^4+y^4)}{(x^2+y^2)^a+^1}[/tex]And this is continuous for all (x,y)==/==0

but i don't see where 3/2 falls out of there.
 
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