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 $$ f(x, y)=5x -10y +y^{3}$$
so we get $$f_x(x, y)=5 , \ \ f_y(x, y)=-10+3 y^2$$
$$\text { Letting } z=f(x, y), \text { the increment of } z \\ \text { at an arbitrary point }(x, y) \text { in the plane is }$$
\begin{aligned}
\Delta z& =f(x+\Delta x, y+\Delta y)-f(x, y)\\
&= 5(x+\Delta x) -10(y+ \Delta y) +(y+\Delta y)^{3} -5x +10y -y^{3}\\
&= 5x+5(\Delta x) -10y-10( \Delta y)+y^3+3y^2 (\Delta y)+3y (\Delta y)^2+(\Delta y)^{3} -5x +10y -y^{3}\\
&= 5(\Delta x) -10 (\Delta y)+3y^2 (\Delta y)+3y (\Delta y)^2+(\Delta y)^{3} \\
&=5(\Delta x)+\left(3 y^{2}-10\right)(\Delta y)+0(\Delta x)+\left(3 y(\Delta y)+(\Delta y)^{2}\right) \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}= 0 \text { and } \varepsilon_{2}=3 y(\Delta y)+(\Delta y)^{2}
\end{aligned}
as
$$ ( \Delta x, \Delta y) \rightarrow(0,0), \varepsilon_{1} \rightarrow 0 \text { and } \varepsilon_{2} \rightarrow 0$$
this implies
$$f \text { is differentiable at every point in the plane }$$
