We are given the function $f(x, y) = xy^3$ and the following conditions:

1. The direction of maximum increase is aligned with the vector $\langle 1, 1 \rangle$.
2. The slope in the direction of maximum increase is $\sqrt{18}$.

Let's analyze this step-by-step.

Step 1: Gradient of the function

The direction of the maximum increase of a function is given by the gradient $\nabla f(x, y)$, which is the vector of partial derivatives with respect to $x$ and $y$. For $f(x, y) = xy^3$, the gradient is:

$\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$

Calculating the partial derivatives:

$\frac{\partial f}{\partial x} = y^3$ $\frac{\partial f}{\partial y} = 3xy^2$

Thus, the gradient is:

$\nabla f(x, y) = \langle y^3, 3xy^2 \rangle$

Step 2: Condition on the direction of the gradient

We are told that the direction of maximum increase is aligned with the vector $\langle 1, 1 \rangle$. This implies that the gradient vector $\nabla f(x, y)$ must be proportional to $\langle 1, 1 \rangle$. In other words, there exists a scalar $\lambda$ such that:

$\nabla f(x, y) = \lambda \langle 1, 1 \rangle$

This gives us the system of equations:

$y^3 = \lambda$ $3xy^2 = \lambda$

Equating the two expressions for $\lambda$:

$y^3 = 3xy^2$

Factoring out $y^2$ (assuming $y \neq 0$):

$y^2(y - 3x) = 0$

So, $y = 0$ or $y = 3x$.

Case 1: $y = 0$

Substituting $y = 0$ into the partial derivatives gives $\nabla f(x, 0) = \langle 0, 0 \rangle$, which is not aligned with $\langle 1, 1 \rangle$. Hence, $y = 0$ does not provide any solutions.

Case 2: $y = 3x$

Substitute $y = 3x$ into one of the equations for $\lambda$:

$y^3 = (3x)^3 = 27x^3$

Thus, $\lambda = 27x^3$.

Step 3: Condition on the slope

The slope in the direction of maximum increase is given by the magnitude of the gradient. We are told that this slope is $\sqrt{18}$. The magnitude of the gradient is:

$|\nabla f(x, y)| = \sqrt{\left( y^3 \right)^2 + \left( 3xy^2 \right)^2}$

Substituting $y = 3x$ into this:

$|\nabla f(x, 3x)| = \sqrt{(27x^3)^2 + (9x^3)^2}$ $|\nabla f(x, 3x)| = \sqrt{729x^6 + 81x^6} = \sqrt{810x^6} = \sqrt{810} |x^3|$ $|\nabla f(x, 3x)| = 9\sqrt{10} |x^3|$

We are told that this equals $\sqrt{18}$, so:

$9\sqrt{10} |x^3| = \sqrt{18}$

Solving for $x$:

$9\sqrt{10} |x^3| = \sqrt{18}$ [ |x^3