If $\lvert f(x)\rvert\leq \lvert x\rvert^2$, then $f$ is differentiable at $0$ 
Let $f:\Bbb{R}^2\to\Bbb{R}$ be a function such that $\lvert f(x)\rvert\leq \lvert x\rvert^2$. Show that $f$ is differentiable at $0$.

My solution:
We want to show that $$\lim_{h\to 0}\dfrac{f(h)-f(0)}{ h}$$ exists. By assumption $f(0)=0$ and $$\dfrac{\lvert f(h)\rvert}{\lvert h\rvert}\leq\dfrac{\lvert h\rvert^2}{\lvert h\rvert}=\lvert h\rvert$$. So by the $\epsilon-\delta$ definition the result follows.
Is this correct?
 A: Correct, except, the definition of the derivative at $x=0$ is defined by
$$
\lim_{h\to 0}\frac{f(h)-f(0)}{h}
$$
and not by
$$
\lim_{h\to 0}\frac{f(h)-f(0)}{|h|}
$$
So, rigorously speaking, in order to show that $f'(0)=0$, you simply need to prove that
$$
\lim_{h\to 0}\frac{f(h)-f(0)}{h}=0,
$$
which is equivalent to
$$
\lim_{h\to 0}\frac{|f(h)-f(0)|}{|h|}=0,
$$
and as
$$
\frac{|f(h)-f(0)|}{|h|}=\frac{|f(h)|}{|h|}\le \frac{|h|^2}{|h|}=|h|,
$$
and since $\lim_{h\to 0}|h|=0$, then also $\lim_{h\to 0}\dfrac{|f(h)-f(0)|}{|h|}=0$.
Note also that $|f(0)|\le 0^2$, and hence $f(0)=0$.
A: Your proof appears wrong because you're using the wrong definition.
A function $f$ is differntiable in a point $x_0$ if $$f(x_0 + h) = f(x_0) + \nabla f (x_0) \cdot h + o(|h|)$$
You have basically to prove that 
$$\lim_{|h| \to 0} \frac{f(x_0 + h) - f(x_0) - \nabla f(x_0) \cdot h}{|h|} = 0$$
(note that $h$ is a vector)
In your case, $x_0 = (0, 0) ; h = (t, k)$ 
$$\lim_{t, k \to 0, 0} \frac{f(t, k) - f(0, 0) - f_x(0, 0) t - f_y(0, 0)k}{\sqrt{t^2 + k^2}} = 0$$
(now $t$ and $k$ are real numbers)
Can you continue from here?
Note that you could also show that $f \in C^1$, as this condition implies differantiablity
