1. State the problem: Find the third derivative $y'''$ of the function $y = x^2 - \sqrt[3]{4x}$. 2. Rewrite the function using exponents: $$y = x^2 - (4x)^{1/3}.$$ 3. Use the chain rule to find the first derivative $y'$. $$y' = \frac{d}{dx}x^2 - \frac{d}{dx}(4x)^{1/3} = 2x - \frac{1}{3}(4x)^{-2/3} \cdot 4.$$ Simplify the second term: $$y' = 2x - \frac{4}{3}(4x)^{-2/3}.$$ 4. Find the second derivative $y''$: $$y'' = \frac{d}{dx} \left( 2x - \frac{4}{3} (4x)^{-2/3} \right) = 2 - \frac{4}{3} \cdot \left(-\frac{2}{3}\right)(4x)^{-5/3} \cdot 4.$$ Calculate: $$y'' = 2 + \frac{8}{9} \cdot 4 (4x)^{-5/3} = 2 + \frac{32}{9}(4x)^{-5/3}.$$ 5. Find the third derivative $y'''$: $$y''' = \frac{d}{dx} \left( 2 + \frac{32}{9} (4x)^{-5/3} \right) = 0 + \frac{32}{9} \cdot \left(-\frac{5}{3}\right)(4x)^{-8/3} \cdot 4.$$ Calculate: $$y''' = -\frac{160}{27} \cdot 4 (4x)^{-8/3} = -\frac{640}{27} (4x)^{-8/3}.$$ 6. Simplify the final answer if desired: $$\boxed{y''' = -\frac{640}{27} (4x)^{-8/3}}.$$ This is the third derivative of the given function.