Prove that if $|f(x,y)|\leq x^2+y^2$ for all $(x,y) \in \mathbb{R}^2$ then $f$ is differentiable at $(0,0)$ 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?
 A: This is all very laboured. A function $f:\Bbb R^2\to\Bbb R$ is differentiable at $(0,0)$ with derivative $a$ if $f(v)-f(0,0)=a\cdot v+o(|v|)$
as $v\to0$ where $v$ is a vector, $a\cdot v$ is the dot product of $a$
and $v$, and $|v|$ is the Euclidean length of $v$. Here
$$f(v)=(0,0)\cdot v+O(|v|^2)$$
so $f$ if differentiable with derivative $(0,0)$ at the origin.
A: $|f_x(0,0)|= \left|\displaystyle \lim_{h \to 0}\dfrac{f(h,0) - f(0,0)}{h}\right|=\displaystyle \lim_{ h \to 0}\left|\dfrac{f(h,0)}{h}\right|\le\displaystyle \lim_{h \to 0} \dfrac{h^2+0^2}{|h|}=0$,and the other partial at $(0,0)$ is also $0$, thus the numerator is $0$, hence the fraction is $0$.
Note: I simply continue what you left off, and to complete the proof you would use all the work you started. 
A: As you showed in your question, $f(0,0)=0$, and $$\lim_{(x,y)\rightarrow(0,0)} \frac{f(x,y) - f(0,0)}{\sqrt{x^2 +y^2}}=\lim_{(x,y)\rightarrow(0,0)} \frac{f(x,y) - 0}{\sqrt{x^2 +y^2}} = \lim_{(x,y)\rightarrow(0,0)} \frac{f(x,y)}{\sqrt{x^2 +y^2}}.$$
So it suffices to show the limit on the right exists.
Claim: $\lim_{(x,y)\rightarrow(0,0)} \frac{f(x,y)}{\sqrt{x^2 +y^2}}= 0$.
Proof: Observe that 
$$\left|\frac{f(x,y)}{\sqrt{x^2 +y^2}}\right| \leq \frac{x^2+y^2}{\sqrt{x^2 +y^2}}  = \sqrt{x^2+y^2}.$$
Hence $\frac{f(x,y)}{\sqrt{x^2 +y^2}}\rightarrow 0$ as $(x,y)\rightarrow 0$. 
