Answer
as
$$ ( \Delta x, \Delta y) \rightarrow(0,0), \varepsilon_{1} \rightarrow 0 \text { and } \varepsilon_{2} \rightarrow 0$$
this implies
$$f \text { is differentiable at every point in the plane }$$
Work Step by Step
Given $$ (x, y)=x^{2}-2 x+y $$
so we get $$f_x(x, y)=2x-2, \ \ f_y(x, y)=1$$
$$\text { Letting } z=f(x, y), \text { the increment of } z \text { at an arbitrary point }(x, y) \text { in the plane is }$$
\begin{aligned}
\Delta z& =f(x+\Delta x, y+\Delta y)-f(x, y)\\
&= (x+\Delta x)^{2}-2 (x+\Delta x)+(y+\Delta y)\\
&=\left(x^{2}+2 x(\Delta x)+(\Delta x)^{2}-2 x-2(\Delta x)+y+(\Delta y)\right)-\left(x^{2}-2 x+y\right) \\ &=2 x(\Delta x)+(\Delta x)^{2}-2(\Delta x)+(\Delta y)\\
&=(2 x-2) \Delta x+\Delta y+\Delta x(\Delta x)+0(\Delta y)
\\ &=f_{x}(x, y) \Delta x+f_{y}(x, y) \Delta y+\varepsilon_{1} \Delta x+\varepsilon_{2} \Delta y \\
& \text { where } \varepsilon_{1}=\Delta x \text { and } \varepsilon_{2}=0
\end{aligned}
as
$$ ( \Delta x, \Delta y) \rightarrow(0,0), \varepsilon_{1} \rightarrow 0 \text { and } \varepsilon_{2} \rightarrow 0$$
this implies
$$f \text { is differentiable at every point in the plane }$$
