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To solve the given problem, we need to use the Chain Rule to compute the partial derivatives of \( M \) with respect to \( u \) and \( v \). Let's walk through this step by step.

### Given:
\[
M = xe^y - z^2
\]
\[
x = 2uv, \quad y = u - v, \quad z = u + v
\]
with \( u = 3 \) and \( v = -1 \).

### 1. Compute \( \frac{\partial M}{\partial u} \)

Using the Chain Rule, we can express \( \frac{\partial M}{\partial u} \) in terms of the partial derivatives of \( M \) with respect to \( x \), \( y \), and \( z \), and the partials of \( x \), \( y \), and \( z \) with respect to \( u \).

\[
\frac{\partial M}{\partial u} = \frac{\partial M}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial M}{\partial y} \cdot \frac{\partial y}{\partial u} + \frac{\partial M}{\partial z} \cdot \frac{\partial z}{\partial u}
\]

Let's compute each term:

- \( \frac{\partial M}{\partial x} = e^y \)
- \( \frac{\partial M}{\partial y} = xe^y \)
- \( \frac{\partial M}{\partial z} = -2z \)

Now for the partials of \( x \), \( y \), and \( z \) with respect to \( u \):

- \( \frac{\partial x}{\partial u} = 2v \)
- \( \frac{\partial y}{\partial u} = 1 \)
- \( \frac{\partial z}{\partial u} = 1 \)

Thus, the expression for \( \frac{\partial M}{\partial u} \) becomes:
\[
\frac{\partial M}{\partial u} = e^y \cdot 2v + xe^y \cdot 1 + (-2z) \cdot 1
\]

### 2. Compute \( \frac{\partial M}{\partial v} \)

Similarly, for \( \frac{\partial M}{\partial v} \), we apply the Chain Rule:

\[
\frac{\partial M}{\partial v} = \frac{\partial M}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial M}{\partial y} \cdot \frac{\partial y}{\partial v} + \frac{\partial M}{\partial z} \cdot \frac{\partial z}{\partial v}
\]

- \( \frac{\partial x}{\partial v} = 2u \)
- \( \frac{\partial y}{\partial v} = -1 \)
- \( \frac{\partial z}{\partial v} = 1 \)

So:
\[
\frac{\partial M}{\partial v} = e^y \cdot 2u + xe^y \cdot (-1) + (-2z) \cdot 1
\]

### Substituting \( u = 3 \) and \( v = -1 \)

- \( x = 2uv = 2(3)(-1) = -6 \)
- \( y = u - v = 3 - (-1) = 4 \)
- \( z = u + v = 3 + (-1) = 2 \)
- \( e^y = e^4 \)

Now we substitute these values into the two equations.

#### For \( \frac{\partial M}{\partial u} \):
\[
\frac{\partial M}{\partial u} = e^4 \cdot 2(-1) + (-6)e^4 \cdot 1 + (-2)(2)
\]
\[
\frac{\partial M}{\partial u} = -2e^4 - 6e^4 - 4 = -8e^4 - 4
\]

#### For \( \frac{\partial M}{\partial v} \):
\[
\frac{\partial M}{\partial v} = e^4 \cdot 2(3) + (-6)e^4 \cdot (-1) + (-2)(2)
\]
\[
\frac{\partial M}{\partial v} = 6e^4 + 6e^4 - 4 = 12e^4 - 4
\]

### Final Answer:

\[
\frac{\partial M}{\partial u} = -8e^4 - 4
\]
\[
\frac{\partial M}{\partial v} = 12e^4 - 4
\]

Would you like further explanation or details on any step?

Here are some related questions for you to explore:

1. How does the Chain Rule work in multivariable calculus?
2. What are the applications of partial derivatives in real-world problems?
3. How do you compute partial derivatives in more complex expressions?
4. How does implicit differentiation differ from using the Chain Rule?
5. What are some other techniques for finding total derivatives?

**Tip:** In problems involving chain rule, always break down the derivative into smaller parts by computing each derivative step-by-step for each variable and expression.

Use the Chain Rule to find the indicated partial derivatives. M = xe^y - z^2, x = 2uv, y = u - v, z = u + v; when u = 3, v = -1

This problem demonstrates how to apply the Chain Rule to compute partial derivatives of M = xe^y - z^2 where x, y, and z depend on u and v. The solution includes detailed steps for calculating ∂M/∂u and ∂M/∂v when u = 3 and v = -1.

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