Why did $h^4$ become $h^3h^4$? Why did $4$ become $2$?
In general, remember that for $a$ and $b$ positive, $\sqrt[4]{ab} = \sqrt[4]{a}\sqrt[4]{b}$, and that $(a^r)^{s} = a^{rs}$. Added: If $g$, $h$, and $r$ are positive, then you can rewrite what you have as: $\sqrt[4]{\frac{g^3h^4}{4r^{14}}} = \left( g^3\times h^4 \times 4^{-1} \times r^{-14}\right)^{1/4},$ and then using the laws of exponents you get $\begin{align*} \left(g^3\times h^4\times 4^{-1} \times r^{-14}\right)^{1/4} &= g^{3/4}\times h^{4/4} \times 4^{-1/4}\times r^{-14/4}\\ &= g^{3/4}\times h \times 4^{-1/4}\times r^{-3}\times r^{-2/4} \\ &= \frac{h\sqrt[4]{g^3}}{r^34^{1/4}r^{1/2}}\\ &= \frac{h\sqrt[4]{g^3}}{r^3(2^2)^{1/4}r^{1/2}}\\ &=\frac{h\sqrt[4]{g^3}}{r^3 2^{1/2}r^{1/2}}\\ &= \frac{h\sqrt[4]{g^3}}{r^3\sqrt{2r}}. \end{align*}$
If $r$ is not known to be positive, then you shoudl replace the $r^3$ and the $r$ in the last step with $|r|^3$ and $|r|$. $g$ must be positive for the original expression to be sensible; if $h$ is not known to be positive, then you should replace the $h$ at the end with $|h|$.