The points on the curve $\frac{x^2}{9}+\frac{y^2}{64}= 1$ at which the tangents are parallel to the x-axis are :

(0, ±8)

$\frac{x^2}{9}+\frac{y^2}{64}=1.$

$\text{Differentiate implicitly:}$

$\frac{2x}{9}+\frac{2y}{64}\frac{dy}{dx}=0.$

$\frac{dy}{dx}=-\frac{64x}{9y}.$

$\text{For tangents parallel to x-axis:}$

$\frac{dy}{dx}=0.$

$-\frac{64x}{9y}=0.$

$x=0.$

$\text{Substitute in ellipse equation:}$

$\frac{y^2}{64}=1.$

$y=\pm8.$

$\text{Points }=(0,8),(0,-8).$