Find the value of $h^{\prime }\left(e\right)$ where $h\left(x\right)={\left(3x\right)}^{\frac{1}{3x}}$, and $e$ is the base of the natural logarithm.

$h^{\prime }\left(e\right)=-\frac{{\left(3e\right)}^{\frac{1}{3e}}\ln 3}{3e^2}$

$h^{\prime }\left(e\right)=\frac{{\left(3e\right)}^{\frac{1}{3e}}\ln 3}{3e^2}$

$h^{\prime }\left(e\right)=-\frac{{\left(3e\right)}^{\frac{1}{3e}}\ln 3}{e^2}$

$h^{\prime }\left(e\right)=\frac{{\left(3e\right)}^{-\frac{1}{3e}}\ln 3}{e^2}$