1. **State the problem:** Solve the differential equation $$(3x^3 - 3x + 1) \frac{dy}{dx} = (3x^2 - 1) y.$$\n\n2. **Rewrite the equation:** We want to express it in the form $$\frac{dy}{dx} = \frac{(3x^2 - 1)}{(3x^3 - 3x + 1)} y.$$\n\n3. **Identify the type:** This is a first-order linear differential equation of the form $$\frac{dy}{dx} = P(x) y,$$ where $$P(x) = \frac{3x^2 - 1}{3x^3 - 3x + 1}.$$\n\n4. **Solve by separation of variables:** We can write $$\frac{1}{y} dy = \frac{3x^2 - 1}{3x^3 - 3x + 1} dx.$$\n\n5. **Integrate both sides:** $$\int \frac{1}{y} dy = \int \frac{3x^2 - 1}{3x^3 - 3x + 1} dx.$$\n\n6. **Left side integral:** $$\int \frac{1}{y} dy = \ln|y| + C_1.$$\n\n7. **Right side integral:** Notice the denominator $$3x^3 - 3x + 1$$ and numerator $$3x^2 - 1$$. The derivative of the denominator is $$9x^2 - 3$$, which is $$3(3x^2 - 1)$$, exactly 3 times the numerator.\n\nSo, $$\frac{3x^2 - 1}{3x^3 - 3x + 1} = \frac{1}{3} \cdot \frac{9x^2 - 3}{3x^3 - 3x + 1} = \frac{1}{3} \cdot \frac{d}{dx}(\ln|3x^3 - 3x + 1|).$$\n\n8. **Therefore,** $$\int \frac{3x^2 - 1}{3x^3 - 3x + 1} dx = \frac{1}{3} \ln|3x^3 - 3x + 1| + C_2.$$\n\n9. **Combine constants and write general solution:**\n$$\ln|y| = \frac{1}{3} \ln|3x^3 - 3x + 1| + C,$$ where $$C = C_2 - C_1.$$\n\n10. **Exponentiate both sides:**\n$$|y| = e^C |3x^3 - 3x + 1|^{1/3}.$$\n\n11. **Write final solution:**\n$$y = C' (3x^3 - 3x + 1)^{1/3},$$ where $$C' = \pm e^C$$ is an arbitrary constant.\n\nThis is the implicit solution without rearranging further as requested.