Answer
$(3,+\infty)$
Work Step by Step
$h(x)=\displaystyle \frac{1}{\sqrt[4]{x-3}}$ which has an even indexed root in the denominator.
First, $x-3\geq 0$, in order for the root to be defined, and
secondly $x\neq 3$, in order for the denominator not to be zero.
Domain: $\left\{\begin{array}{ll}
x\geq 3 & and\\
x\neq 3 &
\end{array}\right.$ ,
is the interval $(3,+\infty)$
