To find the second-order partial derivative of the function $y e^{xy}$ with respect to $x$, we follow these steps:

1. First-order partial derivative with respect to $x$

The given function is: $f(x, y) = y e^{xy}$

Differentiate this with respect to $x$, treating $y$ as a constant: $\frac{\partial}{\partial x} \left( y e^{xy} \right)$

Use the product rule, where $u(x) = y$ and $v(x) = e^{xy}$: $\frac{\partial}{\partial x} \left( y e^{xy} \right) = y \frac{\partial}{\partial x} \left( e^{xy} \right)$

Now, differentiate $e^{xy}$ with respect to $x$: $\frac{\partial}{\partial x} \left( e^{xy} \right) = y e^{xy}$

So the first-order partial derivative becomes: $\frac{\partial}{\partial x} \left( y e^{xy} \right) = y^2 e^{xy}$

2. Second-order partial derivative with respect to $x$

Now, take the derivative of $y^2 e^{xy}$ with respect to $x$: $\frac{\partial}{\partial x} \left( y^2 e^{xy} \right)$

Again, use the product rule, where $u(x) = y^2$ and $v(x) = e^{xy}$: $\frac{\partial}{\partial x} \left( y^2 e^{xy} \right) = y^2 \frac{\partial}{\partial x} \left( e^{xy} \right)$

Since $\frac{\partial}{\partial x} \left( e^{xy} \right) = y e^{xy}$, we get: $\frac{\partial^2}{\partial x^2} \left( y e^{xy} \right) = y^3 e^{xy}$

Thus, the second-order partial derivative of $y e^{xy}$ with respect to $x$ is: $\frac{\partial^2}{\partial x^2} \left( y e^{xy} \right) = y^3 e^{xy}$

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Do you want further details or have any questions? Here are some related questions for deeper understanding:

1. What is the first-order partial derivative of $e^{xy}$ with respect to $y$?
2. How do you find mixed partial derivatives of a function involving exponential terms?
3. Can you generalize the process for finding higher-order derivatives of $e^{xy}$?
4. What is the second-order mixed partial derivative of $y e^{xy}$ with respect to $x$ and $y$?
5. How does the product rule apply when differentiating exponential functions?

Tip: When differentiating exponential functions like $e^{xy}$, always remember to use the chain rule for any non-trivial exponent terms.