To find the gradient of the function $f(x, y) = x e^{xy}$, we need to compute the partial derivatives of $f$ with respect to both $x$ and $y$.

1. Partial derivative with respect to $x$:

We apply the product rule since $f(x, y) = x \cdot e^{xy}$.

$\frac{\partial}{\partial x} [x \cdot e^{xy}] = \frac{\partial}{\partial x} x \cdot e^{xy} + x \cdot \frac{\partial}{\partial x} e^{xy}$

The derivative of $x$ with respect to $x$ is 1, and the derivative of $e^{xy}$ with respect to $x$ is:

$\frac{\partial}{\partial x} e^{xy} = y e^{xy}$

Thus, we have:

$\frac{\partial f}{\partial x} = e^{xy} + x \cdot y e^{xy} = e^{xy}(1 + xy)$

2. Partial derivative with respect to $y$:

Now, we differentiate $f(x, y) = x e^{xy}$ with respect to $y$. Here, $x$ is treated as a constant, and the derivative of $e^{xy}$ with respect to $y$ is:

$\frac{\partial}{\partial y} e^{xy} = x e^{xy}$

Thus:

$\frac{\partial f}{\partial y} = x^2 e^{xy}$

Gradient of $f(x, y)$:

The gradient of $f(x, y)$ is the vector formed by the partial derivatives with respect to $x$ and $y$:

$\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( e^{xy}(1 + xy), x^2 e^{xy} \right)$

Let me know if you need more details or have any questions!

Here are five related questions:

1. How would you find the critical points of $f(x, y) = x e^{xy}$?
2. What is the Hessian matrix for $f(x, y) = x e^{xy}$?
3. How does the gradient help in identifying the direction of steepest ascent?
4. What happens to the gradient when $x = 0$ or $y = 0$?
5. How would you compute the directional derivative of $f(x, y)$ in a given direction?

Tip: The gradient of a function at any point points in the direction of the steepest increase of the function at that point.