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