If `x = cos theta and y = sin^(3) theta,` then `|(yd ^(2)y)/(dx ^(2))+((dy)/(dx))^(2)|at theta=(pi)/(2)` is:

Answer

To solve the problem, we need to find the expression \( | \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 | \) at \( \theta = \frac{\pi}{2} \) given that \( x = \cos \theta \) and \( y = \sin^3 \theta \).

### Step 1: Find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \)

Given:
- \( x = \cos \theta \)
- \( y = \sin^3 \theta \)

Differentiating both with respect to \( \theta \):
\[ \frac{dx}{d\theta} = -\sin \theta \]
\[ \frac{dy}{d\theta} = 3\sin^2 \theta \cos \theta \]

### Step 2: Find \( \frac{dy}{dx} \)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{3\sin^2 \theta \cos \theta}{-\sin \theta} = -3\sin \theta \cos \theta \]

### Step 3: Find \( \frac{d^2y}{dx^2} \)

To find \( \frac{d^2y}{dx^2} \), we differentiate \( \frac{dy}{dx} \) with respect to \( \theta \):
\[ \frac{d}{d\theta} \left( -3\sin \theta \cos \theta \right) = -3 \left( \cos^2 \theta - \sin^2 \theta \right) \]

Now, we apply the chain rule again:
\[ \frac{d^2y}{dx^2} = \frac{d}{d\theta} \left( -3(\cos^2 \theta - \sin^2 \theta) \right) \cdot \frac{1}{\frac{dx}{d\theta}} \]
Substituting \( \frac{dx}{d\theta} \):
\[ \frac{d^2y}{dx^2} = -3(\cos^2 \theta - \sin^2 \theta) \cdot \frac{1}{-\sin \theta} = \frac{3(\cos^2 \theta - \sin^2 \theta)}{\sin \theta} \]

### Step 4: Evaluate at \( \theta = \frac{\pi}{2} \)

At \( \theta = \frac{\pi}{2} \):
- \( \sin \frac{\pi}{2} = 1 \)
- \( \cos \frac{\pi}{2} = 0 \)

Thus:
\[ \frac{dy}{dx} = -3(1)(0) = 0 \]
\[ \frac{d^2y}{dx^2} = \frac{3(0^2 - 1^2)}{1} = -3 \]

### Step 5: Substitute into the expression

Now we substitute into the expression:
\[ \left| \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \right| = \left| -3 + 0^2 \right| = \left| -3 \right| = 3 \]

### Final Answer

The final answer is:
\[ \boxed{3} \]