 Stationary points with real cube-root factors

Where $q(x)\ne 0$, the product rule and chain rule give the displayed real derivative. Its stationary points therefore satisfy $3q+(x-a)q^{\prime }=0$. The denominator uses the square of the real cube root, which stays positive for negative $q$. Points with $q=0$ must be checked in the original function: a simple zero of $q$ with $x\ne a$ creates a vertical tangent, not a stationary point. If the prefactor also vanishes, a cancellation can change differentiability and must be analyzed separately.