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To find the minima and maxima of the function \( f(x, y) = xy + \frac{2}{x} + \frac{4}{y} \), we will first compute the partial derivatives with respect to \(x\) and \(y\), set them equal to zero, and then solve for the critical points. Finally, we will determine the nature of these critical points using the second derivative test.

### Step 1: Find the partial derivatives
1. **Partial derivative with respect to \(x\):**
   \[
   f_x(x, y) = \frac{\partial}{\partial x} \left( xy + \frac{2}{x} + \frac{4}{y} \right)
   \]
   Differentiating term by term:
   \[
   f_x(x, y) = y - \frac{2}{x^2}
   \]

2. **Partial derivative with respect to \(y\):**
   \[
   f_y(x, y) = \frac{\partial}{\partial y} \left( xy + \frac{2}{x} + \frac{4}{y} \right)
   \]
   Differentiating term by term:
   \[
   f_y(x, y) = x - \frac{4}{y^2}
   \]

### Step 2: Set the partial derivatives equal to zero
To find the critical points, we set \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \):

1. **Setting \( f_x(x, y) = 0 \):**
   \[
   y - \frac{2}{x^2} = 0
   \]
   \[
   y = \frac{2}{x^2}
   \]

2. **Setting \( f_y(x, y) = 0 \):**
   \[
   x - \frac{4}{y^2} = 0
   \]
   \[
   x = \frac{4}{y^2}
   \]

### Step 3: Solve the system of equations
Now we have two equations:
\[
y = \frac{2}{x^2}
\]
\[
x = \frac{4}{y^2}
\]

Substituting \( y = \frac{2}{x^2} \) into \( x = \frac{4}{y^2} \):

\[
x = \frac{4}{\left(\frac{2}{x^2}\right)^2} = \frac{4x^4}{4} = x^4
\]
\[
x^4 = x
\]
This implies:
\[
x(x^3 - 1) = 0
\]
So, \( x = 0 \) or \( x = 1 \) or \( x = -1 \).

- **For \( x = 0 \):** Substituting into \( y = \frac{2}{x^2} \) is undefined, so this is not a valid solution.
- **For \( x = 1 \):** \( y = \frac{2}{1^2} = 2 \).
- **For \( x = -1 \):** \( y = \frac{2}{(-1)^2} = 2 \).

Thus, the critical points are \( (1, 2) \) and \( (-1, 2) \).

### Step 4: Second derivative test
To determine the nature of the critical points, we compute the second partial derivatives:

1. **Second partial derivatives:**
   \[
   f_{xx} = \frac{\partial}{\partial x} \left( y - \frac{2}{x^2} \right) = \frac{4}{x^3}
   \]
   \[
   f_{yy} = \frac{\partial}{\partial y} \left( x - \frac{4}{y^2} \right) = \frac{8}{y^3}
   \]
   \[
   f_{xy} = \frac{\partial}{\partial y} \left( y - \frac{2}{x^2} \right) = 1
   \]

2. **Compute the discriminant:**
   The discriminant \( D \) is given by:
   \[
   D = f_{xx} f_{yy} - (f_{xy})^2
   \]
   - At \( (1, 2) \):
     \[
     f_{xx} = 4, \quad f_{yy} = 1, \quad f_{xy} = 1
     \]
     \[
     D = (4)(1) - (1)^2 = 4 - 1 = 3 > 0
     \]
     Since \( D > 0 \) and \( f_{xx} > 0 \), \( (1, 2) \) is a local minimum.

   - At \( (-1, 2) \):
     \[
     f_{xx} = -4, \quad f_{yy} = 1, \quad f_{xy} = 1
     \]
     \[
     D = (-4)(1) - (1)^2 = -4 - 1 = -5 < 0
     \]
     Since \( D < 0 \), \( (-1, 2) \) is a saddle point.

### Conclusion:
- The function \( f(x, y) \) has a local minimum at \( (1, 2) \).
- The function \( f(x, y) \) has a saddle point at \( (-1, 2) \).

Would you like more details on any specific part of this solution, or do you have any other questions?

Here are five related questions you might find interesting:
1. How can the method used above be extended to functions of three variables?
2. What is the geometric interpretation of a saddle point?
3. How can one use the Hessian matrix for more complex functions?
4. Can we determine global minima or maxima for this function?
5. How does the choice of function domain affect the critical points?

**Tip:** When solving for critical points in multivariable functions, always double-check the domain and ensure that the points you find are valid within the context of the problem.

Learn how to find the minima and maxima of a multivariable function using partial derivatives and the second derivative test. This detailed solution covers key concepts in multivariable calculus, including partial derivatives, critical points, and the second derivative test. Suitable for advanced college-level students and anyone interested in multivariable calculus problems.

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