Find the critical point(s) of the function.

Then use the second derivative test to classify the nature of each point, if possible.
Finally, determine the relative extrema of the function

f(x,y)= 3x^2 - 3e^5y^2

There is a lot of work to this problem, so I'll talk you through the steps, but the work must be done by yourself

first you need to find derivatives of fx and fy and then find when they go to zero. Once you figure out what those values you are for fxx and fyy, youll want to plug those values back into the original equation in order to find the z value. Then, youll want to figure out fxx and fyy to "test to classify the nature of each point."
After all that work is said and done,
d=fxx*fyy-fxy^2
if d is less than 0 than its a saddle point
if fxx<o and d>0 its a maxima
if fxx>0 and d>0 its a local minima

3x^2 - 3e^5y^2

To find the critical point(s) of the function f(x, y) = 3x^2 - 3e^(5y^2), we will take the partial derivatives with respect to x and y and set them equal to zero.

The partial derivative with respect to x (denoted as ∂f/∂x) is found by differentiating f(x, y) with respect to x while treating y as a constant:
∂f/∂x = 6x

The partial derivative with respect to y (denoted as ∂f/∂y) is found by differentiating f(x, y) with respect to y while treating x as a constant:
∂f/∂y = -30ye^(5y^2)

Setting ∂f/∂x and ∂f/∂y equal to zero, we have:

6x = 0, which implies x = 0

-30ye^(5y^2) = 0, which gives two possibilities:
1. -30y = 0, which implies y = 0
2. e^(5y^2) = 0, which has no real solutions

Therefore, the critical points of the function are at (0, 0).

Next, we need to classify the nature of the critical point(s) using the second derivative test. For this, we will find the second partial derivatives ∂^2f/∂x^2, ∂^2f/∂x∂y, and ∂^2f/∂y^2.

∂^2f/∂x^2 = 6
∂^2f/∂x∂y = 0
∂^2f/∂y^2 = -150ye^(5y^2) + 150y^2e^(5y^2)

To apply the second derivative test, evaluate these second partial derivatives at the critical point (0, 0).

∂^2f/∂x^2 evaluated at (0, 0) is 6.
∂^2f/∂x∂y evaluated at (0, 0) is 0.
∂^2f/∂y^2 evaluated at (0, 0) is -150(0)e^(5(0)^2) + 150(0)^2e^(5(0)^2) = 0.

By the second derivative test, we determine the nature of each critical point:

If ∂^2f/∂x^2 > 0 and ∂^2f/∂x∂y > 0, then the point is a local minimum.
If ∂^2f/∂x^2 < 0 and ∂^2f/∂x∂y < 0, then the point is a local maximum.
If ∂^2f/∂x^2 > 0 and ∂^2f/∂x∂y < 0 or ∂^2f/∂x^2 < 0 and ∂^2f/∂x∂y > 0, then the point is a saddle point.

In this case, ∂^2f/∂x^2 = 6 > 0 and ∂^2f/∂x∂y = 0, so the critical point (0, 0) is a saddle point.

Finally, to determine the relative extrema of the function, we need to compare the values of f(x, y) at the critical point(s) and the values of f(x, y) at other points.

Substituting (0, 0) into f(x, y), we get:
f(0, 0) = 3(0)^2 - 3e^(5(0)^2) = 0 - 3 = -3

Therefore, the function has a relative minimum at the critical point (0, 0), with f(0, 0) = -3.