I am a little confused on this definition/expression in my textbook . I know how to take a partial derivative, but am confused on what the $ε$ means and stands for.
Definition: If $z = f(x,y)$, then $f$ is differentiable at $(a,b)$ if $\Delta z $ can be expressed in the form $$ \Delta z = f_x (a,b) \Delta x + f_y (a,b) \Delta y + ε_1 \Delta x + ε_2 \Delta y $$
Where $ε_1,ε_2 \to 0$ as $( \Delta x , \Delta y) \to (0,0)$
I think it is best for you to understand differentiability of a univariate function first. We say that a $1$-input function $f$ (whose domain could be for instance a subset of $\mathbb{R}$ or $\mathbb{C}$) is differentiable at a point $x$ (in its domain) iff $f(x+Δx) ∈ f(x) + a·Δx + o(Δx)$ as $Δx → 0$ for some constant $a$ (which may depend on $x$). Here "$o(Δx)$" denotes a set of values that has eventually insignificant magnitude compared to $Δx$ (as $Δx → 0$). This captures the idea that $f$ is close to linear around $x$, since it equivalently asserts that for any real $ε > 0$ we have $f(x+Δx) ∈ f(x) + [a-ε,a+ε]·Δx$ eventually as $Δx → 0$, which can be stated informally as "the graph of $f$ can be squeezed between lines with slope arbitrarily close to $a$ on both sides for points close enough to $x$". $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $
Note that this definition of differentiability of $f$ at $x$ is equivalent to $\lfrac{f(x+Δx)-f(x)}{Δx} ∈ a + o(1)$ as $Δx → 0$ for some constant $a$, which is a trivial restatement of $\lim_{Δx→0} \lfrac{f(x+Δx)-f(x)}{Δx} = a$ for some constant $a$. However, this viewpoint naturally generalizes to multiple-input functions.
The correct intuition is that differentiability of a multivariable function $f$ at a point $v$ means that there is a linear approximation to $f$ around $v$. In your case $f$ has two inputs, so differentiability of $f$ at a point $(x,y)$ means that $f(x+Δx,y+Δy) ∈ f(x,y) + a·Δx + b·Δy + o(Δx) + o(Δy)$ as $Δx,Δy → 0$ for some constants $a,b$ (which may depend on $(x,y)$). As before, "$o(Δx)$" denotes a set of values that is eventually insignificant compared to $Δx$ (as $Δx,Δy → 0$), and similarly for "$o(Δy)$". Observe that $o(Δx) + o(Δy) = o(1)·Δx + o(1)·Δy$, which is the reason your textbook expresses it as $ε_1·Δx + ε_2·Δy$ for some $ε_1,ε_2 → 0$ as $Δx,Δy → 0$. Note also that there is no real need for two epsilons, because $o(1)·Δx + o(1)·Δy = o(1)·(|Δx|+|Δy|)$, so we can alternatively express it as $ε·(|Δx|+|Δy|)$ for some $ε → 0$ as $Δx,Δy → 0$.
Furthermore, since $|t|+|u| = \sqrt{t^2}+\sqrt{u^2} ∈ [1,2]·\sqrt{t^2+u^2}$ for any reals $t,u$, it is also equivalent to expressing it as $ε·\sqrt{Δx^2+Δy^2}$ for some $ε → 0$ as $Δx,Δy → 0$. This gives the vector form of differentiability. A multivariable real function $f : \mathbb{R}^n → \mathbb{R}$ is differentiable at a point $v ∈ \mathbb{R}^n$ iff there is a linear map $J : \mathbb{R}^n → \mathbb{R}$ such that $\lfrac{f(v+h)-f(v)-J(h)}{|h|} ≈ 0$ as $h ∈ \mathbb{R}^n → 0$.
And if $f$ is differentiable, then we can find $a,b$ as follows. $f(x+Δx,y) ∈ f(x,y) + a·Δx + o(Δx)$ as $Δx → 0$, or equivalently $\lfrac{f(x+Δx,y)-f(x,y)}{Δx} ∈ a + o(1)$ as $Δx → 0$, which implies $\lfrac{∂(f(x,y))}{∂x} = a$. Likewise, $\lfrac{∂(f(x,y))}{∂y} = b$. But note that $f$ may not be differentiable at $(x,y)$ even if $f$ is continuous and both $\lfrac{∂(f(x,y))}{∂x}$ and $\lfrac{∂(f(x,y))}{∂y}$ are well-defined there. (Exercise: construct a counter-example.)