Differentiate implicitly to find $\frac{d y}{d x}$. Then, find the slope of the curve at the given point.
\[
x^{2}-3 y^{2}=-4 ; \quad(\sqrt{5}, \sqrt{3})
\]
\[
\frac{d y}{d x}=
\]

Step 1 ：Given the equation \(x^{2}-3 y^{2}=-4\), we can differentiate both sides with respect to \(x\).

Step 2 ：Applying the power rule, we get \(2x - 3 \cdot 2y \cdot \frac{dy}{dx} = 0\).

Step 3 ：Rearranging the equation, we get \(\frac{dy}{dx} = \frac{2x}{6y}\).

Step 4 ：Simplifying the equation, we get \(\frac{dy}{dx} = \frac{x}{3y}\).

Step 5 ：Substitute the given point \((\sqrt{5}, \sqrt{3})\) into the equation, we get \(\frac{dy}{dx} = \frac{\sqrt{5}}{3\sqrt{3}}\).

Step 6 ：Simplify the equation, we get \(\frac{dy}{dx} = \frac{\sqrt{15}}{9}\).

Step 7 ：So, the slope of the curve at the given point \((\sqrt{5}, \sqrt{3})\) is \(\boxed{\frac{\sqrt{15}}{9}}\).