Answer
$$
f(x, y, z)=\ln \left|8xy+5y z-x^{3}\right|
$$
$$
\begin{aligned}
f_{x}(x, y, z)&= \frac{8y-3x^{2}}{8xy+5y z-x^{3}}
\end{aligned}
$$
$$
\begin{aligned}
f_{y}(x, y, z) &=\frac{8x+5z}{8xy+5y z-x^{3}}
\end{aligned}
$$
$$
\begin{aligned}
f_{z}(x, y, z) &=\frac{5y}{8xy+5y z-x^{3}}
\end{aligned}
$$
$$
\begin{aligned} f_{y z}(x, y, z) &=\frac{\partial}{\partial y}(\frac{5y}{8xy+5y z-x^{3}})\\
&=\frac{-5 x^{3}}{\left(8 x y+5 y z-x^{3}\right)^{2}} \end{aligned}
$$
Work Step by Step
$$
f(x, y, z)=\ln \left|8xy+5y z-x^{3}\right|
$$
find $f_{x}(x, y, z).$
Using the formula of a natural logarithm function and the chain rule, treat $y$ and $z$ constants.
$$
\begin{aligned}
f_{x}(x, y, z)&= \frac{1}{8xy+5y z-x^{3}}. \frac{\partial}{\partial x} (8xy+5y z-x^{3}) \\
&=\frac{8y-3x^{2}}{8xy+5y z-x^{3}}
\end{aligned}
$$
find $f_{y}(x, y, z).$
Using the formula of a natural logarithm function and the chain rule, treat $x$ and $z$ constants.
$$
\begin{aligned}
f_{y}(x, y, z) &=\frac{1}{8xy+5y z-x^{3}}. \frac{\partial}{\partial y} (8xy+5y z-x^{3}) \\
&=\frac{8x+5z}{8xy+5y z-x^{3}}
\end{aligned}
$$
find $f_{z}(x, y, z).$
Using the formula of a natural logarithm function and the chain rule, treat $x$ and $y$ constants.
$$
\begin{aligned}
f_{z}(x, y, z) &=\frac{1}{8xy+5y z-x^{3}}. \frac{\partial}{\partial z} (8xy+5y z-x^{3}) \\
&=\frac{5y}{8xy+5y z-x^{3}}
\end{aligned}
$$
find $f_{yz}(x, y, z).$
The derivative of a function $f_{z}(x, y, z)$ relative $y$ and treat $x$ and $z$ constants. By using the quotient rule and the chain rule we have :
$$
\begin{aligned} f_{y z}(x, y, z) &=\frac{\partial}{\partial y}(\frac{5y}{8xy+5y z-x^{3}})\\
&=\frac{\left(8 x y+5 y z-x^{3}\right) \cdot 5-(8 x+5 z) \cdot 5 y}{\left(8 x y+5 y z-x^{3}\right)^{2}} \\ &=\frac{-5 x^{3}}{\left(8 x y+5 y z-x^{3}\right)^{2}}. \end{aligned}
$$
