My attempt: 

We want

> 
> $$\lim\limits\_{(x,y)\to(0,0)} \frac{|f(x,y)-L\_{(0,0)}(x,y)|}{\sqrt{x^2+y^2}}=0 \tag1$$
> 
> 
> 

Setting up for a squeeze/sandwich theorem we have

$$\left|\frac{|f(x,y)-L\_{(0,0)}(x,y)|}{\sqrt{x^2+y^2}}-0\right| $$

By triangle inequality:

$$\leq \frac{|f(x,y)|+ |L\_{(0,0)}(x,y)|}{\sqrt{x^2+y^2}}$$

Also from our assumption $f(0,0)\leq0^2+0^2=0$

$$\leq \frac{0+|L\_{(0,0)}(x,y)|}{\sqrt{x^2+y^2}}$$

Expanding linearization and using our assumption again:

$$=\frac{|f(0,0)+f\_x(0,0)x+f\_y(0,0)y|}{\sqrt{x^2+y^2}}\leq \frac{|0+ f\_x(0,0)x+f\_y(0,0)y|}{\sqrt{x^2+y^2}}$$

This is where I'm stuck, how do I proceed further to get a limit $\to$ 0?  

Or is there a better approach to this problem?

EDIT: tried to find partial derivatives using definition  

(since the limit above would hold true if partials at $(0,0)$ is $0$)

Claim:

> 
> $$f\_x(0,0) = \lim\limits\_{h\to0} \frac{f(0+h,0)-f(0,0)}{h} = 0$$
> 
> 
> 

Proof:
$$\left|\frac{f(0+h,0)-f(0,0)}{h}-0\right|\leq\frac{|f(0+h,0)|+|f(0,0)|}{|h|}\leq\frac{|h^2|+|0|}{|h|}=|h|\to0$$

Hence by squeeze (and symmetry) both $f\_x(0,0) = f\_y(0,0) = 0$ 

Therefore, equation the limit $(1)$ goes to $0$ and thus $f$ is differentiable at $(0,0)$ 

Would this be a complete answer?