Showing that $f(x,y)\sim f(x_0,y_0)+\frac{\partial f}{\partial x}\Delta x + \frac{\partial f}{\partial y}\Delta y$

Suppose $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is differentiable at $(x_0,y_0)$. I want to show that

$$f(x,y) = f(x_0,y_0)+\frac{\partial f}{\partial x}\Delta x + \frac{\partial f}{\partial y}\Delta y +\epsilon (x,y)$$

where $e\rightarrow 0$ as $(x,y)\rightarrow (x_0,y_0)$.

The idea I have in mind is that of

$$f(x,y)\sim f(x_0,y) + \frac{\partial f}{\partial x}\Delta x\sim f(x_0,y_0)+\frac{\partial f}{\partial x}\Delta x + \frac{\partial f}{\partial y}\Delta y$$

The last approximation can be proven easily by regarding $f$ as a function of $y$ alone: let $f_{x_0}(y)=f(x_0,y)$ and note that $$f_{x_0}(y)=f_{x_0}(y_0)+\frac{df}{dy}\Delta y +\epsilon _y\Delta y$$ where $$\epsilon _y =\Big( \frac{f_{x_0}(y)-f_{x_0}(y_0)}{\Delta y}-\frac{df}{dy}\Big) \rightarrow 0 \ \ \text{as} \ \ \Delta y\rightarrow 0$$

Yet I can't prove the first approximation.

I would appreciate any help.

Any differentiable function satisfies it by definition of differentiability, since it implies the existence of a vector $v$ (it can be seen as a linear functional) such that $$ \lim_{(x,y)\rightarrow (x_0,y_0)} \dfrac{f(x,y)-f(x_0,y_0)-\left(v_1 (x-x_0)+v_2(y-y_0)\right)}{|x-x_0|+|y-y_0|}=0 $$ After that, it's easy to see that partial derivatives is the components of the vector $v$, since $$0 = \lim_{t \rightarrow 0}\dfrac{f((x_0,y_0)+t e_i)-f(x_0,y_0)-t v_i}{t},$$ for $i=1$ and $i=2$, where $e_1=(1,0)$ and $e_2=(0,1)$. There is no need of being a big brain here, for more details see article. On the other hand, a function with continuous partial derivatives is differentiable, see the proof, and that is what you were trying to do in your last paragraph. All this things is basically stated in Stewart.

Here, my only concern is about your $\epsilon(x,y)$ error. It needs to be a term of the form $$\epsilon(x,y) = (|x−x_0|+|y−y_0|)g(x,y),$$ with $g(x,y)$ converging to zero as $(x,y)\rightarrow(x_0,y_0)$. After that, it's easy to see that what you are wanting to show is the definition of differentability.