Answer
$$
f(x, y, z)=\ln \left|x^{2}-5 x z^{2}+y^{4}\right|
$$
$$
\begin{aligned}
f_{x}(x, y, z)&= \frac{2 x-5 z^{2}}{x^{2}-5 x z^{2}+y^{4}}
\end{aligned}
$$
$$
\begin{aligned}
f_{y}(x, y, z) &=\frac{4 y^{3}}{x^{2}-5 x z^{2}+y^{4}}
\end{aligned}
$$
$$
\begin{aligned}
f_{z}(x, y, z) &=\frac{-10 x z}{x^{2}-5 x z^{2}+y^{4}}
\end{aligned}
$$
$$
\begin{aligned}
f_{yz}(x, y, z) &=\frac{40 x y^{3} z}{\left(x^{2}-5 x z^{2}+y^{4}\right)^{2}}.
\end{aligned}
$$
Work Step by Step
$$
f(x, y, z)=\ln \left|x^{2}-5 x z^{2}+y^{4}\right|
$$
find $f_{x}(x, y, z).$
Using the formula of a natural logarithm function and the chain rule, treat $y$ and $z$ constants.
$$
\begin{aligned}
f_{x}(x, y, z)&= \frac{1}{x^{2}-5 x z^{2}+y^{4}}. \frac{\partial}{\partial x} (x^{2}-5 x z^{2}+y^{4}) \\
&=\frac{2 x-5 z^{2}}{x^{2}-5 x z^{2}+y^{4}}
\end{aligned}
$$
find $f_{y}(x, y, z).$
Using the formula of a natural logarithm function and the chain rule, treat $x$ and $z$ constants.
$$
\begin{aligned}
f_{y}(x, y, z) &=\frac{1}{x^{2}-5 x z^{2}+y^{4}}. \frac{\partial}{\partial y} (x^{2}-5 x z^{2}+y^{4}) \\
&=\frac{4 y^{3}}{x^{2}-5 x z^{2}+y^{4}}
\end{aligned}
$$
find $f_{z}(x, y, z).$
Using the formula of a natural logarithm function and the chain rule, treat $x$ and $y$ constants.
$$
\begin{aligned}
f_{z}(x, y, z) &=\frac{1}{x^{2}-5 x z^{2}+y^{4}}. \frac{\partial}{\partial z} (x^{2}-5 x z^{2}+y^{4}) \\
&=\frac{-10 x z}{x^{2}-5 x z^{2}+y^{4}}
\end{aligned}
$$
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 :
$$
\begin{aligned}
f_{yz}(x, y, z) &=\frac{\partial}{\partial y}(\frac{-10 x z}{x^{2}-5 x z^{2}+y^{4}})\\
&= \frac{4 y^{3}(10 z x)}{\left(x^{2}-5 x z^{2}+y^{4}\right)^{2}} \\
&=\frac{40 x y^{3} z}{\left(x^{2}-5 x z^{2}+y^{4}\right)^{2}}.
\end{aligned}
$$
