Answer
A constant function of two or three variables can be differentiable everywhere.
Work Step by Step
According to definition 13.4.1, a function is said to be differentiable at $\left(x_{0}, y_{0}\right)$ if partial derivatives exist:
\[
\left.\lim _{(\Delta x, \Delta y) \rightarrow(0,0)} \frac{\Delta f-f_{x}\left(x_{0}, y_{0}\right) \Delta x-f_{y}\left(x_{0},\right.}{\sqrt{(\Delta y)^{2}+(\Delta x)^{2}}}\right) \Delta y=0
\]
Evaluating the limit if it's equal to zero, our function is differentiable everywhere
\[
0=0 \\ \& f_{y}\left(x_{0}, y_{0}\right)=f_{x}(x, y)
\]
Also
\[
\begin{array}{l}
\Delta f=f\left(x_{0}+\Delta x, y_{0}+\Delta y\right)-f\left(x_{0}, y_{0}\right) \\
0=k-k=\Delta f
\end{array}
\]
Thus,
\[
\lim _{(\Delta x, \Delta y) \rightarrow(0,0)} \frac{\Delta f-f_{x}\left(x_{0}, y_{0}\right) \Delta x-f_{y}\left(x_{0}, y_{0}\right) \Delta y}{\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}}=\lim _{(\Delta x, \Delta y) \rightarrow(0,0)} \frac{0-0 \Delta x-0 \Delta y}{\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}}
\]
\[
=0
\]
Thus, a constant function of two or three variables can be differentiable everywhere.
