1. **State the problem:** We need to find the derivative $\frac{dy}{dx}$ at the point $(1,1)$ for the curve given by the implicit equation $$2y^3 - 3y + 3 = 2x^3.$$\n\n2. **Formula and rules:** Since $y$ is implicitly defined as a function of $x$, we use implicit differentiation. Differentiate both sides of the equation with respect to $x$, remembering to apply the chain rule for terms involving $y$.\n\n3. **Differentiate both sides:**\n$$\frac{d}{dx}(2y^3) - \frac{d}{dx}(3y) + \frac{d}{dx}(3) = \frac{d}{dx}(2x^3)$$\nUsing the chain rule:\n$$2 \cdot 3y^2 \frac{dy}{dx} - 3 \frac{dy}{dx} + 0 = 6x^2$$\nSimplify:\n$$6y^2 \frac{dy}{dx} - 3 \frac{dy}{dx} = 6x^2$$\n\n4. **Factor out $\frac{dy}{dx}$:**\n$$\left(6y^2 - 3\right) \frac{dy}{dx} = 6x^2$$\n\n5. **Solve for $\frac{dy}{dx}$:**\n$$\frac{dy}{dx} = \frac{6x^2}{6y^2 - 3}$$\n\n6. **Evaluate at the point $(1,1)$:**\nSubstitute $x=1$ and $y=1$:\n$$\frac{dy}{dx}\bigg|_{(1,1)} = \frac{6 \cdot 1^2}{6 \cdot 1^2 - 3} = \frac{6}{6 - 3} = \frac{6}{3} = 2.$$\n\n**Final answer:** The slope of the curve at the point $(1,1)$ is $$\boxed{2}.$$