Answer
$-\frac{3}{4}$
Work Step by Step
Step 1. Letting $f^{-1}(-\frac{3}{5})=sin^{-1}(-\frac{3}{5})=t$, we have $sin(t)=-\frac{3}{5}$ and $t$ in quadrant IV.
Step 2. Form a right triangle with sides $y=3, r=5, x=\sqrt {r^2-y^2}=4$ with $|t|$ facing $y$.
Step 3. We have $h(f^{-1}(-\frac{3}{5}))=tan(t)=-\frac{3}{4}$