#### Answer

$(a)$. $$\frac{dy}{dx}=\frac{-3}{2y}$$
$(b)$. The slope of the tangent line to the curve is $m=\frac{-3\sqrt{5}}{10}$

#### Work Step by Step

$(a)$ Use implicit differentiation to determine $\frac{dy}{dx}$ for $y^2+3x=8$
Taking the derivative implicitly we get:
$$2y\frac{dy}{dx}+3=0$$
solve for $\frac{dy}{dx}$
$$\frac{dy}{dx}=\frac{-3}{2y}$$
$(b)$. Find the slope of the tangent line to the curve at $(1,\sqrt{5})$
We plug in the point $(1,\sqrt{5})$ into the derivative from part $(a)$. Hence:
$$\frac{dy}{dx}=\frac{-3}{2\sqrt{5}}=\frac{-3\sqrt{5}}{10}$$