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