I have been taught that $F$ is differentiable at a point, say $(a,b)$, if there exists a real valued continuous function $W(h,k)$ such that it tends to zero as $(h,k)$ approaches $(0,0)$ satisfying:
$$F(a+h,b+k)=F(a,b) + h\cdot D_1F(a,b))+k \cdot D_2F((a,b))+\sqrt(h^2+k^2)\cdot W(h,k)$$
Or if there exists two functions $J_1(h,k),J_2(h,k)$ (both tends to zero as $(h,k)$ approaches zero) such that: $F(a+h,b+k)=F(a,b)+h\cdot D_1F(a,b))+k\cdot D_2F((a,b)) + h\cdot J_1+k J_2$
I would like to know how these two conditions are equivalent.
Accepting the second definition I was able to arrive at the first; but not the other way around. Any help is appreciated...
Partial result for the original question:
The first condition, i.e. the existence of a continuous function $W$ with $W(0,0)=0$ satisfying $$F(a+h,b+k)=F(a,b) + h\cdot D_1F(a,b)+k \cdot D_2F(a,b)+(h^2+k^2)\cdot W(h,k)$$ for all $h,k$, is far too restrictive to be a condition of differentiability of $F$ at $(a,b)$. Even the nicest functions may not be "differentiable" under this definition.
Consider $F(a,b)=a^2+b^2$. We want to check the "differentiability" of $F$ at $(a,b)=(0,0)$. The partial derivatives are zero at this point. We want to show the nonexistence of a continuous function $W$ with $W(0,0)=0$ satisfying $$F(h,k)=(h^2+k^2)\cdot W(h,k).$$ Note that $F(h,k)=h^2+k^2$. So if $W$ exists, then $$(h^2+k^2)(1-W(h,k))=0$$ for all $h,k$. Choose any pair of $h,k$ such that they are not simultaneously zero. For such $h,k$, we have $h^2+k^2\not=0$ and so we must have $1-W(h,k)=0$ for all pairs of $h,k$ not simultaneously zero. Take limit of this expression $(h,k)\to(0,0)$ and we have $$\lim\limits_{(h,k)\to(0,0)}1-W(h,k)=0.$$ But we also have that $$\lim\limits_{(h,k)\to(0,0)}1-W(h,k)=1-\lim\limits_{(h,k)\to(0,0)}W(h,k)=1-0=1.$$ So no such function $W$ exists. Even some polynomials can be non-differentiable under this condition.
A suggested correction to the first condition is the following:
The function $W$ would be required to be continuous only at $(0,0)$. The equation would be corrected to $$F(a+h,b+k)=F(a,b) + h\cdot D_1F(a,b)+k \cdot D_2F(a,b)+\sqrt{h^2+k^2}\cdot W(h,k).$$ The rationale of this correction is as follows: differentiability is commonly defined by that $F$ is differentiable at $(a,b)$ if the following limit is true: $$\lim\limits_{(h,k)\to(0,0)}\frac{F(a+h,b+k)-F(a,b) - h\cdot D_1F(a,b)-k \cdot D_2F(a,b)}{\sqrt{h^2+k^2}}=0.$$ Then, we could define $$W(h,k)=\begin{cases}\frac{F(a+h,b+k)-F(a,b) - h\cdot D_1F(a,b)-k \cdot D_2F(a,b)}{\sqrt{h^2+k^2}}& \text{if }(h,k)\not=(0,0)\\0&\text{if }(h,k)=(0,0)\end{cases}.$$ This $W$ would be continuous at $(0,0)$, but not necessarily at any other points. That's why we relax the continuity of $W$ to that it is continuous only at $(0,0)$.
Edit: I should stress that the function $W$ you want to find is continuous only if $F$ is continuous. This is important, because if $W$ is continuous, the formula implies that $F$ is continuous. If $F$ happens to be differentiable only at a point but is discontinuous elsewhere, then $W$ must be discontinuous except at $(0,0)$
Now that we corrected the definition. To prove the equivalence of the two definitions, you prove that each of them is equivalent to that the following limit holds:
$$\lim\limits_{(h,k)\to(0,0)}\frac{F(a+h,b+k)-F(a,b) - h\cdot D_1F(a,b)-k \cdot D_2F(a,b)}{\sqrt{h^2+k^2}}=0.$$
As a note, this limit is equivalent to that $$\lim\limits_{(h,k)\to(0,0)}\frac{F(a+h,b+k)-F(a,b) - h\cdot D_1F(a,b)-k \cdot D_2F(a,b)}{\lvert h\rvert+\lvert k\rvert}=0.$$ This is due to norm equivalence between $\sqrt{h^2+k^2}$ and $\lvert h\rvert+\lvert k\rvert$:
$$\frac{1}{\sqrt2}(\lvert h\rvert+\lvert k\rvert)\le\sqrt{h^2+k^2}\le\lvert h\rvert+\lvert k\rvert$$
As an exercise, prove that the above two limits hold simultaneously or not hold simultaneously, using this provided inequality.
Showing that your first definition (the one with $W$) is equivalent to my first limit is almost trivial, because the whole expression in my limit is just the function $W$.
Showing the equivalence of your second definition and my second limit is slightly more messy. We first prove that my limit implies your second definition.
Define $$J(h,k)=\begin{cases}\frac{F(a+h,b+k)-F(a,b) - h\cdot D_1F(a,b)-k \cdot D_2F(a,b)}{\lvert h\rvert+\lvert k\rvert}& \text{if }(h,k)\not=(0,0)\\0&\text{if }(h,k)=(0,0)\end{cases}.$$ Then $J_1,J_2$ are given by $$J_1(h,k)=\begin{cases}J(h,k)& \text{if }h\ge0\\-J(h,k)&\text{if }h\lt0,\end{cases}$$ $$J_2(h,k)=\begin{cases}J(h,k)& \text{if }k\ge0\\-J(h,k)&\text{if }k\lt0.\end{cases}$$ They are so defined so that $(\lvert h\rvert+\lvert k\rvert)J(h,k)=h\cdot J_1(h,k)+k\cdot J_2(h,k)$ for all $h,k$. You should verify that $J_1,J_2$ tends to zero as $(h,k)\to(0,0)$.
To prove your second definition implies my second limit, you just need to prove the limit $$\lim\limits_{(h,k)\to(0,0)}\frac{h\cdot J_1(h,k)+k\cdot J_2(h,k)}{\lvert h\rvert+\lvert k\rvert}=0$$ because $h\cdot J_1(h,k)+k\cdot J_2(h,k)=F(a+h,b+k)-F(a,b)-h\cdot D_1F(a,b)-k\cdot D_2F(a,b)$. You only need to show that $$\lim\limits_{(h,k)\to(0,0)}\frac{h\cdot J_1(h,k)}{\lvert h\rvert+\lvert k\rvert}=0$$ because you can then add the two limits (which are both zero).
If $h\not=0$, then $$\frac{\lvert h\cdot J_1(h,k)\rvert}{\lvert h\rvert+\lvert k\rvert}\le\frac{\lvert h\cdot J_1(h,k)\rvert}{\lvert h\rvert}=\lvert J_1(h,k)\rvert\to0$$ as $(h,k)\to(0,0)$. So your second definition does imply my second limit.