To find the equation of the tangent line at the given point $(\pi^2, \pi)$ for the curve $3 \sin y + x = y^2$, we will follow these steps:

1. Verify the point lies on the curve:

The point is $(\pi^2, \pi)$. We substitute $x = \pi^2$ and $y = \pi$ into the given equation:

$3 \sin(\pi) + \pi^2 = \pi^2.$

Since $\sin(\pi) = 0$, this simplifies to:

$0 + \pi^2 = \pi^2,$

which is true, so the point $(\pi^2, \pi)$ does lie on the curve.

2. Implicit Differentiation:

We need to find the slope of the tangent line by differentiating the given equation implicitly:

$3 \sin y + x = y^2.$

Taking the derivative with respect to $x$ on both sides:

$\frac{d}{dx}(3 \sin y) + \frac{d}{dx}(x) = \frac{d}{dx}(y^2),$

which becomes:

$3 \cos y \cdot \frac{dy}{dx} + 1 = 2y \cdot \frac{dy}{dx}.$

Now, solve for $\frac{dy}{dx}$ (which is the slope of the tangent line):

$3 \cos y \cdot \frac{dy}{dx} + 1 = 2y \cdot \frac{dy}{dx}.$

Group the $\frac{dy}{dx}$ terms together:

$\frac{dy}{dx}(3 \cos y - 2y) = -1.$

Thus, the slope is:

$\frac{dy}{dx} = \frac{-1}{3 \cos y - 2y}.$

3. Evaluate the slope at $y = \pi$:

Substitute $y = \pi$ into the slope equation:

$\frac{dy}{dx} = \frac{-1}{3 \cos(\pi) - 2\pi}.$

Since $\cos(\pi) = -1$, this simplifies to:

$\frac{dy}{dx} = \frac{-1}{3(-1) - 2\pi} = \frac{-1}{-3 - 2\pi} = \frac{1}{3 + 2\pi}.$

4. Equation of the Tangent Line:

The tangent line at $(\pi^2, \pi)$ has the form:

$y - y_1 = m(x - x_1),$

where $m = \frac{1}{3 + 2\pi}$, $x_1 = \pi^2$, and $y_1 = \pi$.

Substitute these values into the equation:

$y - \pi = \frac{1}{3 + 2\pi}(x - \pi^2).$

Finally, solving for $y$ in slope-intercept form:

$y = \frac{1}{3 + 2\pi}(x - \pi^2) + \pi.$

This is the equation of the tangent line at $(\pi^2, \pi)$.

Let me know if you'd like further clarifications or details!

Here are five related questions for further practice:

1. How would the equation of the tangent line change if the point were $(4, 2)$ on the same curve?
2. How do you verify that a point lies on a curve in general?
3. What is the geometric meaning of the derivative in the context of tangent lines?
4. Can you find the second derivative $\frac{d^2y}{dx^2}$ at $(\pi^2, \pi)$?
5. How does implicit differentiation differ from explicit differentiation?

Tip: When dealing with implicit functions, always remember to apply the chain rule when differentiating terms that involve the dependent variable!