Answer
$\frac{∂^{3}V}{∂r∂s∂t}=\frac{12st^{2}}{(r+s^{2}+t^{3})^{3}}$
Work Step by Step
Consider the function $V=ln(r+s^{2}+t^{3})$
Let us start differentiating the function with respect to $t$ keeping $r$ and $s$ constant.
$\frac{∂V}{∂t}=\frac{∂}{∂t}[ln(r+s^{2}+t^{3})]$
$=\frac{1}{(r+s^{2}+t^{3})}\frac{∂}{∂t}[(r+s^{2}+t^{3})]$
$=\frac{3t^{2}}{(r+s^{2}+t^{3})}$
Differentiate with respect to $s$ keeping $r$ and $t$ constant.
$\frac{∂^{2}V}{∂s∂t}=-\frac{3t^{2}}{(r+s^{2}+t^{3})^{2}}.2s$ (Apply power rule)
$=-6st^{2}(r+s^{2}+t^{3})^{-2}$
Differentiate with respect to $r$ keeping $s$ and $t$ constant.
$\frac{∂^{3}V}{∂r∂s∂t}=\frac{∂}{∂r}[\frac{∂^{2}}{∂s∂t}(-6st^{2}(r+s^{2}+t^{3})^{-2})]$
$=12st^{2}(r+s^{2}+t^{3})^{-3}$
Hence, $\frac{∂^{3}V}{∂r∂s∂t}=\frac{12st^{2}}{(r+s^{2}+t^{3})^{3}}$
