1. **State the problem:** Find the vertex of the quadratic function $$y = -3x^2 + 4x - 2$$. 2. **Recall the vertex form:** A quadratic function can be written as $$y = a(x-h)^2 + k$$ where $$(h,k)$$ is the vertex. 3. **Given completed square form:** From the user's work, the function is rewritten as $$y = -3\left(x + \frac{4}{6}\right)^2 - 2$$. 4. **Simplify the fraction:** $$\frac{4}{6} = \frac{2}{3}$$, so the function is $$y = -3\left(x + \frac{2}{3}\right)^2 - 2$$. 5. **Identify the vertex:** The vertex is at $$\left(-\frac{2}{3}, -2\right)$$ because the form is $$y = a(x - h)^2 + k$$ and here $$x - h = x + \frac{2}{3}$$ means $$h = -\frac{2}{3}$$. 6. **Interpretation:** The parabola opens downward (since $$a = -3 < 0$$) and the vertex is the maximum point at $$\left(-\frac{2}{3}, -2\right)$$. **Final answer:** The vertex is $$\boxed{\left(-\frac{2}{3}, -2\right)}$$.