#### Answer

\[{h^,}\,\left( 3 \right) = \frac{{13}}{{16}}\]

#### Work Step by Step

\[\begin{gathered}
Let\,\,h\,\left( x \right) = \frac{{f\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\
Find\,\,{h^,}\,\left( x \right)\,\,using\,\,the\,\,quotient\,\,rule \hfill \\
\,\,{\left[ {\frac{u}{v}} \right]^,} = \frac{{v{u^,} - u{v^,}}}{{{v^2}}} \hfill \\
Then \hfill \\
{h^,}\,\left( x \right) = \frac{{g\,\left( x \right){f^,}\,\left( x \right) - f\,\left( x \right){g^,}\,\left( x \right)}}{{\,\,{{\left[ {g\,\left( x \right)} \right]}^2}}} \hfill \\
Then \hfill \\
{h^,}\,\left( 3 \right) = \frac{{g\,\left( 3 \right){f^,}\,\left( 3 \right) - f\,\left( 3 \right){g^,}\,\left( 3 \right)}}{{\,\,{{\left[ {g\,\left( 3 \right)} \right]}^2}}} \hfill \\
Substitute\,\,the\,\,given\,\,values\,\,and\,\,simplify \hfill \\
{h^,}\,\left( 3 \right) = \frac{{\,\left( 4 \right)\,\left( 8 \right) - \,\left( 9 \right)\,\left( 5 \right)}}{{\,{{\left( 4 \right)}^2}}} \hfill \\
{h^,}\,\left( 3 \right) = \frac{{13}}{{16}} \hfill \\
\end{gathered} \]