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To solve this problem, we need to find the partial derivatives \( f_x(x, y) \), \( f_y(x, y) \), \( f_{xy}(x, y) \), and \( f_{yx}(x, y) \) of the function:

\[
f(x, y) = x y e^{1/y}
\]

### Step 1: Calculate \( f_x(x, y) \)
The partial derivative of \( f \) with respect to \( x \) is found by differentiating \( f(x, y) = x y e^{1/y} \) with respect to \( x \), treating \( y \) as a constant.

\[
f_x(x, y) = \frac{\partial}{\partial x} \left( x y e^{1/y} \right) = y e^{1/y}
\]

### Step 2: Calculate \( f_y(x, y) \)
The partial derivative of \( f \) with respect to \( y \) is found by differentiating \( f(x, y) = x y e^{1/y} \) with respect to \( y \), treating \( x \) as a constant.

To differentiate with respect to \( y \), we apply the product rule, as \( f(x, y) = x \cdot (y e^{1/y}) \):

\[
f_y(x, y) = x \cdot \frac{d}{dy}\left(y e^{1/y}\right)
\]

Now, differentiate \( y e^{1/y} \) with respect to \( y \):

\[
\frac{d}{dy}\left(y e^{1/y}\right) = e^{1/y} + y \cdot \frac{d}{dy}\left(e^{1/y}\right)
\]

Since \( \frac{d}{dy}\left(e^{1/y}\right) = e^{1/y} \cdot \left(-\frac{1}{y^2}\right) \):

\[
= e^{1/y} - \frac{e^{1/y}}{y} = e^{1/y} \left(1 - \frac{1}{y}\right)
\]

Thus,

\[
f_y(x, y) = x \cdot e^{1/y} \left(1 - \frac{1}{y}\right) = x e^{1/y} \left(1 - \frac{1}{y}\right)
\]

### Step 3: Calculate \( f_{xy}(x, y) \)
The mixed partial derivative \( f_{xy}(x, y) \) is the partial derivative of \( f_x(x, y) = y e^{1/y} \) with respect to \( y \).

\[
f_{xy}(x, y) = \frac{\partial}{\partial y} \left(y e^{1/y}\right)
\]

As computed earlier:

\[
f_{xy}(x, y) = e^{1/y} \left(1 - \frac{1}{y}\right)
\]

### Step 4: Calculate \( f_{yx}(x, y) \)
Since mixed partial derivatives are usually equal if the function is continuous and well-behaved, we have:

\[
f_{yx}(x, y) = f_{xy}(x, y) = e^{1/y} \left(1 - \frac{1}{y}\right)
\]

### Summary of Results
1. \( f_x(x, y) = y e^{1/y} \)
2. \( f_y(x, y) = x e^{1/y} \left(1 - \frac{1}{y}\right) \)
3. \( f_{xy}(x, y) = e^{1/y} \left(1 - \frac{1}{y}\right) \)
4. \( f_{yx}(x, y) = e^{1/y} \left(1 - \frac{1}{y}\right) \)

Would you like further details or have any questions?

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**Related Questions:**
1. How do we determine if mixed partial derivatives are equal?
2. What is the significance of the exponential function \( e^{1/y} \) in partial derivatives?
3. How does the product rule apply in partial differentiation?
4. What happens if \( y = 0 \) in this function?
5. Can you apply these derivatives to analyze the behavior of \( f(x, y) \) near specific points?

**Tip:** Always check for continuity when assuming mixed partial derivatives are equal.

Find the partial derivatives of the function f(x, y) = xye^{1/y}.

This solution provides step-by-step calculations for finding the partial derivatives of the function f(x, y) = xye^(1/y). The problem includes finding f_x, f_y, f_{xy}, and f_{yx} by applying calculus techniques such as the product rule and exponential function differentiation. Suitable for undergraduate calculus students.

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