#### Answer

1) The point $(0,\pi)$ is on the curve.
2) The tangent line to the curve at $(0,\pi)$ is $y=\pi$.
3) The normal line to the curve at $(0,\pi)$ is $x=0$.

#### Work Step by Step

$$x^2\cos^2y-\sin y=0$$
1) Verify that the point $(0,\pi)$ is on the curve:
We substitute the coordinates of the point $(0,\pi)$ into the formula of the function to see if it equals $0$ or not:
$$0^2\cos^2\pi-\sin\pi=0-0=0$$
Since it equals $0$, the point is on the curve.
2) Find the tangent at $(0,\pi)$:
- Find the derivative of the function with implicit differentiation:
$$2x\cos^2y+x^2(2\cos y)(\cos y)'-(\cos y)y'=0$$ $$2x\cos^2y+2x^2\cos y(-\sin y)y'-(\cos y)y'=0$$ $$2x\cos^2y-(2x^2\sin y\cos y)y'-(\cos y)y'=0$$ $$(2x^2\sin y\cos y)y'+(\cos y)y'=2x\cos^2y$$ $$y'(2x^2\sin y\cos y+\cos y)=2x\cos^2y$$ $$y'=\frac{2x\cos^2y}{2x^2\sin y\cos y+\cos y}$$
- The slope of the tangent line at $(0,\pi)$ is
$$y'=\frac{2\times0\cos^2\pi}{2\times0^2\sin\pi\cos\pi+\cos\pi}=\frac{0}{0-1}=0$$
- The tangent line to the given curve at $(0,\pi)$ is
$$y-\pi=0(x-0)$$ $$y-\pi=0$$ $$y=\pi$$
3) Find the normal line at $(0,\pi)$:
Since the tangent line to the curve at $(0,\pi)$ is the horizontal line $y=\pi$, the normal line at this point, being perpendicular with the tangent, will be a vertical line with the form $x=x_0$.
And as the normal line is at $(0,\pi)$, we have $x_0=0$.
The normal line in need to find, hence, is $$x=0$$