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The given function is \( f(x, y) = x^2 e^{-x - 2y} \), and you are asked to compute the value of \( f_x(0,0) + f_y(0,0) + f_{yy}(0,0) \).

### Step-by-Step Solution:

1. **Compute \( f_x(x, y) \):**
   \[
   f(x, y) = x^2 e^{-x - 2y}
   \]
   To find the partial derivative with respect to \( x \), apply the product and chain rules:
   \[
   f_x(x, y) = 2x \cdot e^{-x - 2y} + x^2 \cdot (-1) e^{-x - 2y}
   \]
   Simplifying:
   \[
   f_x(x, y) = e^{-x - 2y} \left( 2x - x^2 \right)
   \]
   Evaluate at \( (x, y) = (0, 0) \):
   \[
   f_x(0, 0) = e^{-(0) - 2(0)} \cdot (2(0) - 0^2) = 0
   \]

2. **Compute \( f_y(x, y) \):**
   Now, take the partial derivative with respect to \( y \):
   \[
   f_y(x, y) = x^2 \cdot (-2) e^{-x - 2y}
   \]
   Simplify:
   \[
   f_y(x, y) = -2x^2 e^{-x - 2y}
   \]
   Evaluate at \( (x, y) = (0, 0) \):
   \[
   f_y(0, 0) = -2(0^2) \cdot e^{-(0) - 2(0)} = 0
   \]

3. **Compute \( f_{yy}(x, y) \):**
   First, recall that:
   \[
   f_y(x, y) = -2x^2 e^{-x - 2y}
   \]
   Now take the partial derivative of this expression with respect to \( y \):
   \[
   f_{yy}(x, y) = -2x^2 \cdot (-2) e^{-x - 2y}
   \]
   Simplify:
   \[
   f_{yy}(x, y) = 4x^2 e^{-x - 2y}
   \]
   Evaluate at \( (x, y) = (0, 0) \):
   \[
   f_{yy}(0, 0) = 4(0^2) e^{-(0) - 2(0)} = 0
   \]

### Final Result:
\[
f_x(0, 0) + f_y(0, 0) + f_{yy}(0, 0) = 0 + 0 + 0 = 0
\]

Therefore, the value of the expression is **0**.

Would you like more details on any step or have further questions?

Here are some related questions for deeper understanding:
1. How would the result change if the function involved a different exponential term?
2. What are the rules for finding partial derivatives of products involving exponentials?
3. How can you use Taylor series to approximate the value of functions around a point like \( (0,0) \)?
4. Can you find the second mixed partial derivative \( f_{xy}(x,y) \) for this function?
5. What does the result \( f_x(0, 0) + f_y(0, 0) + f_{yy}(0, 0) = 0 \) indicate about the behavior of the function at the origin?

**Tip**: When differentiating products, be sure to apply the product rule carefully, especially with functions involving exponentials.

Let f : ℝ^2 → ℝ be defined by f(x, y) = x^2e^(-x - 2y), for every (x, y). The value of f_x(0,0) + f_y(0,0) + f_yy(0,0) is: 0, Undefined, -2, or 2?

Learn how to compute the partial derivatives f_x(0,0), f_y(0,0), and f_yy(0,0) for the function f(x, y) = x^2 e^(-x - 2y). This solution uses key calculus concepts such as the product rule and chain rule to find the result, suitable for students in university-level calculus.

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