Final answer:
To find the value of y" at x = 0, we differentiate the given equation twice with respect to y. The second derivative gives us a value of 0 for y" at x = 0.
Step-by-step explanation:
To find the value of y" at the point where x = 0, we need to differentiate the given equation twice with respect to y. Let's start by finding the first derivative of xy + 6e^y = 6e. Taking the derivative of the left side of the equation, we get x(dy/dx) + 6e^y(dy/dx) = 0. Since x = 0, the first term on the left side becomes 0 and we are left with 6e^y(dy/dx) = 0.
Next, we take the second derivative with respect to y, which gives us 6e^y(dy/dx)^2 + 6e^y(d^2y/dx^2) = 0.
Again, substituting x = 0, the first term becomes 0 and we are left with 6e^y(d^2y/dx^2) = 0. Therefore, at the point where x = 0, the value of y" is 0.