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The image provides the function \( z = e^{-4x} y^2 + \frac{e^{-4y}}{x^2} \) and asks for the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).

### Step-by-step solution:

#### 1. **Partial Derivative with respect to \( x \):**
For \( z = e^{-4x} y^2 + \frac{e^{-4y}}{x^2} \), we treat \( y \) as a constant.

- The derivative of \( e^{-4x} y^2 \) with respect to \( x \):
  \[
  \frac{\partial}{\partial x}(e^{-4x} y^2) = y^2 \cdot \frac{d}{dx}(e^{-4x}) = -4 y^2 e^{-4x}
  \]

- The derivative of \( \frac{e^{-4y}}{x^2} \) with respect to \( x \):
  \[
  \frac{\partial}{\partial x}\left( \frac{e^{-4y}}{x^2} \right) = e^{-4y} \cdot \frac{d}{dx}\left( \frac{1}{x^2} \right) = e^{-4y} \cdot \left( -\frac{2}{x^3} \right) = -\frac{2e^{-4y}}{x^3}
  \]

Thus, the partial derivative \( \frac{\partial z}{\partial x} \) is:
\[
\frac{\partial z}{\partial x} = -4y^2 e^{-4x} - \frac{2e^{-4y}}{x^3}
\]

#### 2. **Partial Derivative with respect to \( y \):**
For the same function \( z = e^{-4x} y^2 + \frac{e^{-4y}}{x^2} \), we now treat \( x \) as a constant.

- The derivative of \( e^{-4x} y^2 \) with respect to \( y \):
  \[
  \frac{\partial}{\partial y}(e^{-4x} y^2) = e^{-4x} \cdot \frac{d}{dy}(y^2) = 2y e^{-4x}
  \]

- The derivative of \( \frac{e^{-4y}}{x^2} \) with respect to \( y \):
  \[
  \frac{\partial}{\partial y}\left( \frac{e^{-4y}}{x^2} \right) = \frac{1}{x^2} \cdot \frac{d}{dy}(e^{-4y}) = \frac{1}{x^2} \cdot (-4e^{-4y}) = -\frac{4e^{-4y}}{x^2}
  \]

Thus, the partial derivative \( \frac{\partial z}{\partial y} \) is:
\[
\frac{\partial z}{\partial y} = 2y e^{-4x} - \frac{4e^{-4y}}{x^2}
\]

### Final answers:
- \( \frac{\partial z}{\partial x} = -4y^2 e^{-4x} - \frac{2e^{-4y}}{x^3} \)
- \( \frac{\partial z}{\partial y} = 2y e^{-4x} - \frac{4e^{-4y}}{x^2} \)

Would you like more details or have any further questions?

Here are 5 related questions:
1. How do we interpret the partial derivatives geometrically?
2. What happens if \( x = 0 \) in the original function \( z \)?
3. Can we apply this process to a function with more complex exponents, like \( z = e^{f(x, y)} \)?
4. How does the chain rule apply when differentiating more complex functions?
5. How would the solution change if the exponents of \( x \) and \( y \) were different?

**Tip:** Always check the domain of the function before differentiating, as values like \( x = 0 \) can cause undefined results in certain terms.

Find the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) for the function \( z = e^{-4x} y^2 + \frac{e^{-4y}}{x^2} \).

This problem demonstrates how to find the partial derivatives of a function involving exponentials and multiple variables. The function in question is \( z = e^{-4x} y^2 + \frac{e^{-4y}}{x^2} \), and the steps include calculating \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) using multivariable calculus techniques.

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