Answer
Function is differentiable at (0,0,0)
Work Step by Step
According to definition 13.4.2, a function is said to be differentiable at $\left(x_{0}, y_{1}, z_{0}\right)$ if partial derivatives exist and
\[
\lim _{(\Delta x, \Delta y, \Delta z) \rightarrow(0,0)} \frac{\Delta f-\int_{x}\left(x_{0}, y_{1}\right) \Delta x-f_{y}\left(x_{0}, y_{0}\right) \Delta y-f_{2}\left(x_{0}, y_{0}\right) \Delta z}{\sqrt{(\Delta x)^{2}+(\Delta y)^{2}+(\Delta z)^{2}}}=0
\]
Thus:
\[
f_{x}(x, y, z)=2 x \\ f_{y}(x, y, z)=2 y \\ f_{z}(x, y, z)=2 z
\]
Also
\[
\begin{array}{r}
\Delta f=f\left(x_{0}+\Delta x, y_{0}+\Delta y, z_{0}+\Delta z\right)-f\left(x_{0}, y_{0}, z_{0}\right) \\
\Delta f=\left(x_{0}+\Delta x\right)^{2}+\left(y_{0}+\Delta y\right)^{2}+\left(z_{0}+\Delta z\right)^{2}-x_{0}^{2}-y_{0}^{2}-z_{0}^{2} \\
\Delta f=(\Delta z)^{2}+(\Delta z)^{2}+(\Delta x)^{2}+(z \Delta z+y \Delta y+x \Delta x)2
\end{array}
\]
Thus
\[
\begin{array}{l}
=\lim _{(\Delta x, \Delta y, \Delta x) \rightarrow(0,0,0)} \frac{\Delta f-f_{x}\left(x_{0}, y_{0}\right) \Delta x-f_{y}\left(x_{0}, y_{0}\right) \Delta y-f_{z}\left(x_{0}, y_{0}\right) \Delta z}{\sqrt{(\Delta y)^{2}+(\Delta x)^{2}}} \\
=\lim _{(\Delta x, \Delta y, \Delta z) \rightarrow(0,0,0)} \frac{(\Delta x)^{2}+(\Delta y)^{2}+(\Delta z)^{2}+2(x \Delta x+y \Delta y+z \Delta z)-2 x \Delta x-2 y \Delta y-2 z \Delta z}{\sqrt{(\Delta x)^{2}+(\Delta y)^{2}+(\Delta z)^{2}}} \sqrt{(\Delta z)^{2}+(\Delta y)^{2}+(\Delta x)^{2}} \\
=\lim _{(\Delta x, \Delta y, \Delta z) \rightarrow(0,0,0)},
\end{array}
\]
$=0$
Thus, it's Differentiable at (0,0,0)
