1. **Problem statement:** Given the function $$y = x^4 - x^3 + 4x - 1,$$ find the second derivative $$\frac{d^2y}{dx^2}$$ and determine the values of $$x$$ for which $$\frac{d^2y}{dx^2} = 0.$$\n\n2. **Recall the formulas:**\n- The first derivative $$\frac{dy}{dx}$$ is the rate of change of $$y$$ with respect to $$x$$.\n- The second derivative $$\frac{d^2y}{dx^2}$$ is the derivative of the first derivative, representing the curvature or concavity of the function.\n- Power rule for differentiation: $$\frac{d}{dx} x^n = n x^{n-1}.$$\n\n3. **Find the first derivative:**\n$$y = x^4 - x^3 + 4x - 1$$\nApply the power rule term by term:\n$$\frac{dy}{dx} = 4x^3 - 3x^2 + 4 - 0 = 4x^3 - 3x^2 + 4.$$\n\n4. **Find the second derivative:**\nDifferentiate $$\frac{dy}{dx}$$ again:\n$$\frac{d^2y}{dx^2} = \frac{d}{dx}(4x^3 - 3x^2 + 4) = 12x^2 - 6x + 0 = 12x^2 - 6x.$$\n\n5. **Set the second derivative equal to zero to find critical points:**\n$$12x^2 - 6x = 0.$$\n\n6. **Factor the equation:**\n$$6x(2x - 1) = 0.$$\n\n7. **Solve for $$x$$:**\nSet each factor equal to zero:\n- $$6x = 0 \Rightarrow x = 0,$$\n- $$2x - 1 = 0 \Rightarrow x = \frac{1}{2}.$$\n\n**Final answer:**\nThe second derivative is $$\frac{d^2y}{dx^2} = 12x^2 - 6x,$$ and the values of $$x$$ for which $$\frac{d^2y}{dx^2} = 0$$ are $$x = 0$$ and $$x = \frac{1}{2}.$$