1. **State the problem:** Find the equation of the line passing through the point $(-4,3)$ and perpendicular to the line given by $$y = -\frac{5}{3}x - 1.$$\n\n2. **Recall the slope of the given line:** The slope-intercept form is $$y = mx + b,$$ where $m$ is the slope. Here, the slope is $$m = -\frac{5}{3}.$$\n\n3. **Find the slope of the perpendicular line:** Lines that are perpendicular have slopes that are negative reciprocals. So, if the original slope is $$m = -\frac{5}{3},$$ the perpendicular slope $$m_\perp$$ is $$m_\perp = -\frac{1}{m} = -\frac{1}{-\frac{5}{3}} = \frac{3}{5}.$$\n\n4. **Use point-slope form to find the equation:** The point-slope form is $$y - y_1 = m(x - x_1),$$ where $(x_1, y_1)$ is the point $(-4,3)$ and $m = \frac{3}{5}$. Substitute these values:\n$$y - 3 = \frac{3}{5}(x - (-4)) = \frac{3}{5}(x + 4).$$\n\n5. **Simplify the equation:**\n$$y - 3 = \frac{3}{5}x + \frac{3}{5} \times 4 = \frac{3}{5}x + \frac{12}{5}.$$\n\n6. **Isolate $y$ to get slope-intercept form:**\n$$y = \frac{3}{5}x + \frac{12}{5} + 3.$$\n\n7. **Convert 3 to a fraction with denominator 5:**\n$$3 = \frac{15}{5},$$ so\n$$y = \frac{3}{5}x + \frac{12}{5} + \frac{15}{5} = \frac{3}{5}x + \frac{27}{5}.$$\n\n**Final answer:** $$y = \frac{3}{5}x + \frac{27}{5}.$$