Answer
True
Work Step by Step
\[
x y z+x^{2}+\ln \left(\frac{y}{z}\right)=f(x, y, z)
\]
1.
\[
\frac{\partial f}{\partial x}=y z+2 x=f_{x}
\]
2.
\[
f_{x y}=\frac{\partial^{2} f}{\partial x y}=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial f}{\partial y}(y ;+2 x)=z
\]
3.
\[
f_{x y z}=\frac{\partial^{3} f}{\partial: y x}=\frac{\partial}{\partial z}\left(\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x} f\right)\right)=\frac{\partial}{\partial z} z=1
\]
4.
\[
\begin{aligned}
f_{x y z x}=& \frac{\partial^{4} f}{\partial x_{2} y x}=\frac{\partial}{\partial x} 1=0 \\
& \therefore {f_{x y z x}=0}
\end{aligned}
\]
To find $f_{2 x x y}$, we will find
1.
\[
f_{z}=\frac{\partial f}{\partial z}=x y-\frac{1}{z}
\]
2.
\[
f_{z x}=\frac{\partial^{2} f}{\partial x z}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial z}\right)=\frac{\partial f}{\partial x}\left(x y-\frac{1}{z}\right)=y
\]
3.
\[
f_{x x x}=\frac{\partial^{3} f}{\partial x x z}=\frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}\left(\frac{\partial}{\partial z} f\right)\right)=\frac{\partial}{\partial x} y=0
\]
4.
\[
\begin{aligned}
f_{x y z x}=& \frac{\partial^{1} f}{\partial y x x z}=\frac{\partial}{\partial y} 0=0 \\
& \therefore\left[f_{z x x y}=0\right.
\end{aligned}
\]
