We are given the function:
\(y = x^3 - x\)

To find the critical points, we set the derivative equal to zero:
\(3x^2 - 1 = 0\)
Solving this gives:
\(x = \pm \frac{1}{\sqrt{3}}\)

Using the second derivative test, we find that:
\(\frac{d^2y}{dx^2} = 6x\)
- For \( x = \frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} > 0 \) (minimum).
- For \( x = -\frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} < 0 \) (maximum).

Thus, the function \( y = x^3 - x \) has a minimum value at:
\(x = \frac{1}{\sqrt{3}} \quad \text{or approximately} \quad x \approx 0.577\)