Suppose $f(\frac{}{3})=4$ and $f^{\text{'}}(\frac{}{3})=-3.$ Let $g(x)=f(x)sin(x)$ and $h(x)=\frac{cos(x)}{f(x)}.$ Find the following.

$($a$)g^{\text{'}}(\frac{}{3})$

$($b$)h^{\text{'}}(\frac{}{3})$

$\frac{3-4\sqrt[\phantom{2}]{3}}{8}$