Critical points are points of a function where the slope is zero (horizontal), and so they could potentially be a maximum or minimum. To find them in this case, let's take the derivative of the function:
g(x) = e-6x^2
g'(x) = e-6x^2 * (-12x)
We want to find the x values that would make this derivative equal to zero. There are no x values that would make the exponential term zero, so the only critical point would occur at:
-12x = 0, so x=0
The derivative is equal to 0 at x=0, so this could be a maximum or a minimum. There are a couple of ways to determine which - let's take the second derivative, since we have to do it for the second part of the question anyway:
g''(x) = e-6x^2 * (-12) + (-12x)*(e-6x^2)*(-12x)
g''(x) = e-6x^2 * (144x2 - 12)
Since we know that there is a critical point at x=0, let's plug in x to determine the concavity. If it comes out to be a positive number, the function is concave up at that point, meaning the function reaches a minimum. If it turns out to be a negative number, it is concave down at that point, meaning the function reaches a maximum.
g''(0) = e0 * (0 - 12) = -12
At x=0, the slope is zero, and the function is concave down, so the function reaches a maximum at x=0.
Inflection points are points where the second derivative is equal to 0, meaning the concavity changes from up to down or vice versa. Looking back at our second derivative expression, again, the exponential term will never be equal to zero, so the only critical points exist wherever:
144x2 - 12 = 0
144x2 = 12
x2 = 12/144 = 1/12
x = +/- √(1/12)
The function has inflection points (where concavity changes) at x = -√(1/12) and x = √(1/12)
I hope this helps!