\( \left. \begin{array} { l }{ \frac { d y } { d x } = 4 x ^ { 3 } + 36 x } \\ { 0 = \frac { d y } { d x } } \end{array} \right. \)

To find the critical points of the function defined by the derivative \( \frac{dy}{dx} = 4x^3 + 36x \), set the derivative equal to zero: \[ 4x^3 + 36x = 0 \] This simplifies to \( 4x(x^2 + 9) = 0 \). The first factor gives us \( x = 0 \) as a solution, while the second factor \( x^2 + 9 = 0 \) has no real solutions since \( x^2 = -9 \) leads to complex numbers. Thus, the only critical point in the real number domain is at \( x = 0 \). Next, we can examine this critical point's nature (minimum or maximum) by evaluating the second derivative. Start by finding the second derivative: \[ \frac{d^2y}{dx^2} = 12x^2 + 36 \] Since \( 12x^2 + 36 > 0 \) for all real numbers \( x \), the function is concave up at \( x = 0 \). Hence, \( x = 0 \) is a local minimum for the function \( y \).