The critical point $\left(\frac{3}{2}, 0\right)$ has been found. Let's summarize the key values:

• First derivatives at the critical point: $f_x\left(\frac{3}{2}, 0\right) = 0, \quad f_y\left(\frac{3}{2}, 0\right) = 0$

• Second derivatives at the critical point: $f_{xx}\left(\frac{3}{2}, 0\right) = 0.2108, \quad f_{yy}\left(\frac{3}{2}, 0\right) = 0.2108$ $f_{xy}\left(\frac{3}{2}, 0\right) = 0$

• Hessian determinant at the critical point: $D = 0.0444$

Classification:

Since the Hessian determinant $D > 0$ and both $f_{xx}$ and $f_{yy}$ are positive, the critical point $\left(\frac{3}{2}, 0\right)$ is a local minimum.

Do you have any questions or would you like more details on this problem?

Here are some related questions you might consider:

1. What if the Hessian determinant was negative—how would we classify the critical point?
2. Can you derive the second derivatives and Hessian determinant manually?
3. How do critical points relate to the global minimum or maximum of a function?
4. What is the significance of the exponential function in optimization problems?
5. How can you visualize the behavior of the function around the critical point?

Tip: Always check the second derivative test when identifying the nature of critical points to avoid incorrect conclusions.