To solve this problem, we start with the equation provided:

$x^2 \tan y = 9$

and are asked to show that

$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}.$

Let's go through the steps to differentiate this equation with respect to $x$.

Step 1: Differentiate Implicitly with Respect to $x$

1. Given: $x^2 \tan y = 9.$

2. Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^2 \tan y) = \frac{d}{dx}(9).$

3. The right side simplifies to zero, as $9$ is a constant: $\frac{d}{dx}(x^2 \tan y) = 0.$

Step 2: Apply the Product Rule

To differentiate $x^2 \tan y$, we use the product rule:

$\frac{d}{dx}(x^2) \cdot \tan y + x^2 \cdot \frac{d}{dx}(\tan y) = 0.$

1. Differentiate $x^2$: $\frac{d}{dx}(x^2) = 2x.$

2. Differentiate $\tan y$ with respect to $x$: $\frac{d}{dx}(\tan y) = \sec^2 y \cdot \frac{dy}{dx}.$

Substitute these derivatives back:

$2x \tan y + x^2 \sec^2 y \cdot \frac{dy}{dx} = 0.$

Step 3: Solve for $\frac{dy}{dx}$

Rearrange the equation to isolate $\frac{dy}{dx}$:

$x^2 \sec^2 y \cdot \frac{dy}{dx} = -2x \tan y.$

$\frac{dy}{dx} = \frac{-2x \tan y}{x^2 \sec^2 y}.$

Step 4: Substitute $\tan y = \frac{9}{x^2}$

From the original equation $x^2 \tan y = 9$, we know:

$\tan y = \frac{9}{x^2}.$

Substitute this into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-2x \cdot \frac{9}{x^2}}{x^2 \sec^2 y}.$

Simplify the numerator:

$\frac{dy}{dx} = \frac{-18x}{x^4 \sec^2 y}.$

Step 5: Express $\sec^2 y$ in Terms of $x$

Since $\tan y = \frac{9}{x^2}$, we have:

$\sec^2 y = 1 + \tan^2 y = 1 + \left(\frac{9}{x^2}\right)^2.$

Simplify this:

$\sec^2 y = 1 + \frac{81}{x^4} = \frac{x^4 + 81}{x^4}.$

Step 6: Substitute $\sec^2 y$ into $\frac{dy}{dx}$

Now, substitute $\sec^2 y = \frac{x^4 + 81}{x^4}$:

$\frac{dy}{dx} = \frac{-18x}{x^4} \cdot \frac{x^4}{x^4 + 81}.$

The $x^4$ terms cancel:

$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}.$

Thus, we have shown the required result:

$\frac{dy}{dx} = \frac{-18x}{x^4 + 81}.$

Would you like further details on any step?

Here are some related questions that might help deepen your understanding:

1. How does implicit differentiation work, and why is it useful in cases like these?
2. Can we use this method to find higher derivatives of $y$ with respect to $x$?
3. What happens to the behavior of $\frac{dy}{dx}$ as $x$ approaches zero?
4. Why do we need to express $\sec^2 y$ in terms of $x$ in this problem?
5. How would the solution change if we had a different trigonometric function, like $\sin y$ or $\cos y$, instead of $\tan y$?

Tip: When working with implicit differentiation, always check if you can express all terms in terms of a single variable to simplify the result.