Finding the Differential $L \in \hom(\mathbb{R}^2, \mathbb{R})$ to show that $f$ is differentiable at the origin 
Problem: Let $f: \mathbb{R}^2 \to \mathbb{R}$ and $\varphi: \mathbb{R} \to \mathbb{R}$ such that $f(x,y)=\varphi(|xy|)$ for all $(x,y) \in \mathbb{R}^2$ Show that:
If $\varphi(0)=0$ and $\exists \delta > 0 : |\varphi(u)| \leq u^2$ for all $|u| < \delta$ then $f$ is differentiable at $(0,0)$


My approach (Updated): I want to show that there exists a linear mapping $L$ from $\mathbb{R}^2 \to \mathbb{R}$ such that $$ \lim_{v \to 0} \frac{f(0+v)-f(0)-L(v)}{|v|}=0  \tag{*}$$
Luckily I have that $f(0)=f(0,0)=\varphi(|0 \cdot 0|)=0$, so the nominator simplifies to $f(v)-L(v)$. What's left to do is to find said linear mapping, which is precisely where I am having difficulty with.
If I say $v:=(v_1,v_2) \in \mathbb{R}^2$ then I can always find $v \in \mathbb{R}^2$ such that $|v_1v_2| < \delta$, it would follow that $f(v)=f(v_1,v_2)=\varphi(|v_1v_2|) \leq (v_1v_2)^2$.
This argument is good enough for a proof because clearly $A:=\lbrace v \in \mathbb{R}^2 : |v| < \delta \text{ and } |v_1v_2| < \delta \rbrace \subset \lbrace v \in \mathbb{R}^2 : |v| < \delta \rbrace $. Thus if I find a $\delta>0$ in the left subset it is more specialized and therefore good enough for a proof.
However if $|v_1v_2| < \delta(v) \implies (v_1v_2)^2 \leq (\delta(v))^2$,  so I have $f(v) -L(v) \leq (\delta(v))^2 - L(v)$, choosing this notation to empathize that $\delta>0$ clearly depends on $v \in \mathbb{R}^2$. I want to construct $L: \mathbb{R}^2 \to \mathbb{R}$ linear such that this distance becomes arbitrary small.
Now I cannot think of many ways to construct a linear mapping from $\mathbb{R}^2 \to \mathbb{R}$, I cannot use any norms because they aren't linear, so I must most likely map $v=(v_1,v_2) \in \mathbb{R}^2$ to a coordinate or some constant. But there must be only one way to do so in order to fulfill (*) because the linear mapping is unique.
(Sorry for this update, what I have written in the last one was ridiculously wrong)
 A: I might be wrong here, because I have very rarely done such proofs, but I believe the existence of partials here is enough, because your $\mathbb{R}^2\rightarrow\mathbb{R}$ map will be basically a multiplication with a row vector whose components are partials at $0$.
To see that, review the definition of the differential and remember that all linear functionals on a coordinate space (whose elements we view as columns) are basically multiplication with rows.
Now, let $g:\mathbb{R}^2\rightarrow\mathbb{R},\ g(x,y)=|xy|$ be a function, then $f=\varphi\circ g$.
Let us calculate the partials of $f$ at $(0,0)$! $$ \left.\frac{\partial f}{\partial x}\right|_{(0,0)}=\left.\frac{d\varphi}{dg(x,y)}\right|_{g(0,0)}\cdot\left.\frac{\partial g}{\partial x}\right|_{(0,0)}. $$ Since $g(0,0)=0,$ what we need to check is $$ \left.\frac{d\varphi}{dx}\right|_{0}=\lim_{\epsilon\rightarrow 0}\frac{\varphi(\epsilon)-\varphi(0)}{\epsilon}. $$We know that $\varphi(0)=0$, and $|\varphi(\epsilon)|\le\epsilon^2$, for small enough $\epsilon$-s.
Using this, $$ \lim_{\epsilon\rightarrow 0}\frac{\varphi(\epsilon)-\varphi(0)}{\epsilon}=\lim_{\epsilon\rightarrow 0}\frac{\varphi(\epsilon)}{\epsilon} $$ and $$ \lim_{\epsilon\rightarrow 0}-\frac{\epsilon^2}{\epsilon}\le\lim_{\epsilon\rightarrow 0}\frac{\varphi(\epsilon)}{\epsilon}\le\lim_{\epsilon\rightarrow 0}\frac{\epsilon^2}{\epsilon}, $$ and since both the left side and the right side converge to $0$, by the policeman principle, or how the hell it is called in english, this is $$ 0\le\lim\frac{\varphi(\epsilon)}{\epsilon}\le0, $$ thus the limit exists and is zero.
Now to check the differentiability of $g(x,y)$, this is a multivariable function, but is symmetric in the variables, thus we only need to check one partial: $$ \left.\frac{\partial g}{\partial x}\right|_{(0,0)}=\left.\frac{\partial |xy|}{\partial x}\right|_{(0,0)}=\lim_{\epsilon\rightarrow 0}\frac{|\epsilon\cdot0|-|0|}{\epsilon}=\lim_{\epsilon\rightarrow 0}\frac{|0|}{\epsilon}=0. $$
All our functions are differentiable at $0$, and $$ \left.\frac{\partial f}{\partial x}\right|_{(0,0)}=0,\ \left.\frac{\partial f}{\partial y}\right|_{(0,0)}=0, $$ so the differential $Df|_{(0,0)}=(0,0).$
A: I don't believe the other answer is correct. You want to show:
$$\lim_{v \rightarrow 0}\frac{f(v) - L(v)}{|v|} = 0$$
For some $L$. As we've seen, $f(0,x) = f(y,0) = \phi(0) = 0$, so the partial derivatives $\frac{\partial}{\partial x}f(0,0)$ and $\frac{\partial}{\partial y}f(0,0)$ exist and equal $0$. This implies that $L$ must be zero, so we need to show:
$$\lim_{v \rightarrow 0}\frac{f(v)}{|v|} = \lim_{(x,y) \rightarrow (0,0)}\frac{\phi(|xy|)}{|(x,y)|} = 0$$
Where $|(x,y)| = \sqrt{x^2 + y^2}$. We know $|x| \leq |(x,y)|$ and $|y| \leq |(x,y)|$. Let's assume $\delta < 1$ (If not just pick a smaller $\delta$). Then if $|(x,y)| < \delta$, $|xy| < \delta^2 < \delta$, so $|\phi(|xy|)| \leq |xy|^{2} \leq |(x,y)|^{4}$. So we get:
$$\frac{|\phi(|xy|)|}{|(x,y)|} \leq \frac{|(x,y)|^{4}}{|(x,y)|} = |(x,y)|^{3} \rightarrow 0$$
And therefore the original limit goes to zero as required. 
