Answer
$\frac{∂^{3}w}{∂z∂y∂x}=\frac{4}{(y+2z)^{3}}$
and
$\frac{∂^{3}w}{∂x^{2}∂y}=0$
Work Step by Step
Consider the function $w=\frac{x}{y+2z}$
Let us start differentiating the function with respect to $x$ keeping $y$ and $z$ constant.
$\frac{∂w}{∂x}=\frac{∂}{∂x}[\frac{x}{y+2z}]$
$=\frac{1}{{y+2z}}$
Differentiate with respect to $y$ keeping $x$ and $z$ constant.
$\frac{∂^{2}w}{∂y∂x}=\frac{∂}{∂y}[\frac{1}{{y+2z}}]$
$=-\frac{1}{(y+2z)^{2}}$
Differentiate with respect to $z$ keeping $x$ and $y$ constant.
$\frac{∂^{3}w}{∂z∂y∂x}=\frac{∂}{∂z}[-\frac{1}{(y+2z)^{2}}]$
$=\frac{2}{(y+2z)^{3}}(2)$
$=\frac{4}{(y+2z)^{3}}$
Thus, $\frac{∂^{3}w}{∂z∂y∂x}=\frac{4}{(y+2z)^{3}}$
Let us start differentiating the function with respect to $y$ keeping $x$ and $z$ constant.
$\frac{∂w}{∂y}=\frac{∂}{∂y}[\frac{x}{y+2z}]$
$=-\frac{x}{{(y+2z})^{2}}$
and
$\frac{∂^{2}w}{∂x∂y}=\frac{∂}{∂x}[-\frac{x}{{(y+2z})^{2}}]$
$=-\frac{1}{{(y+2z})^{2}}$
Again differentiating $\frac{∂^{2}w}{∂x∂y}$ with respect to $x$, we have
Hence, $\frac{∂^{3}w}{∂x^{2}∂y}=0$
