2) The tangent to the curve \( y=x^{3}-\frac{2}{9} \ln x^{4} \) is parallel to \( x- \) axi at \( x=\cdots \) \( \begin{array}{llll}\text { a) } \frac{2}{3} & \text { b) } \pm \frac{2}{3} & \text { c) } \frac{3}{2} & \text { d) } \pm \frac{3}{2}\end{array} \)

To find the points where the tangent to the curve \( y = x^3 - \frac{2}{9} \ln x^4 \) is parallel to the x-axis, we need to determine where the derivative of the function is equal to zero. The derivative \( y' \) represents the slope of the tangent line. First, we differentiate the function \( y \): \[ y = x^3 - \frac{2}{9} \ln x^4 \] Using the properties of logarithms, we can simplify \( \ln x^4 \): \[ \ln x^4 = 4 \ln x \] Thus, we can rewrite \( y \): \[ y = x^3 - \frac{2}{9} \cdot 4 \ln x = x^3 - \frac{8}{9} \ln x \] Now, we differentiate \( y \): \[ y' = \frac{d}{dx}(x^3) - \frac{d}{dx}\left(\frac{8}{9} \ln x\right) \] The derivative of \( x^3 \) is \( 3x^2 \), and the derivative of \( \ln x \) is \( \frac{1}{x} \). Therefore, we have: \[ y' = 3x^2 - \frac{8}{9} \cdot \frac{1}{x} \] To combine the terms, we can express \( y' \) with a common denominator: \[ y' = 3x^2 - \frac{8}{9x} = \frac{27x^3 - 8}{9x} \] Next, we set the derivative equal to zero to find where the tangent is parallel to the x-axis: \[ \frac{27x^3 - 8}{9x} = 0 \] This implies: \[ 27x^3 - 8 = 0 \] Solving for \( x \): \[ 27x^3 = 8 \] \[ x^3 = \frac{8}{27} \] \[ x = \sqrt[3]{\frac{8}{27}} = \frac{2}{3} \] Thus, the value of \( x \) where the tangent to the curve is parallel to the x-axis is: \[ \boxed{\frac{2}{3}} \]