Determine a point on the curve x=cos^3θ,y=sin^3θ,0<=θ<=pi/2 where the tangent is parallel to the curve joining (a,0) and (0,a)?

dx/dθ = -3 cos²θ sinθ
dy/dθ = 3 sin²θ cosθ

Slope of the tangent at (x, y) = dy/dx = (3 sin²θ cosθ)/(-3 cos²θ sinθ) = -tanθ

As the tangent is parallel to the line joining (a, 0) and (0, a), then the slope of the tangent :
-tanθ = (a - 0)/(0 - a)
-tanθ = -1
tanθ = 1
θ = π/4

When θ = π/4 :
x = cos³(π/4), then x = (√2)/4
y = cos³(π/4), then y = (√2)/4

The point on the curve is ((√2)/4, (√2)/4)

Hint:
solve , for θ
dy/dx=-1

　
Slope (a,0) to (0,a) = −1

x = cos³θ
dx/dθ = −3 cos²θ sinθ

y = sin³θ
dy/dθ = 3 sin²θ cosθ

dy/dx = (dy/dθ)/(dx/dθ)
dy/dx = (3 sin²θ cosθ) / (−3 cos²θ sinθ)
dy/dx = −tanθ

So we need to find point(s) where dy/dx = −1
−tanθ = −1
tanθ = 1
sinθ = cosθ = 1/√2 ----> x = y = 1/(2√2)
sinθ = cosθ = −1/√2 ----> x = y = −1/(2√2)

So there are 2 points where tangent is parallel to the line joining (a,0) and (0,a):
(1/(2√2), 1/(2√2)) and (−1/(2√2), −1/(2√2))

https://www.desmos.com/calculator/87uytarqlg

Find the slope between (a , 0) and (0 , a)

(a - 0) / (0 - a) = a/(-a) = -1

Now, derive

x = cos(t)^3
dx/dt = -3 * cos(t)^2 * sin(t)

y = sin(t)^3
dy/dt = 3 * sin(t)^2 * cos(t)

(dy/dt) / (dx/dt) =>
dy/dx =>
(3 * sin(t)^2 * cos(t)) / (-3 * sin(t) * cos(t)^2) =>
-sin(t)/cos(t) =>
-tan(t)

-tan(t) = -1
tan(t) = 1
t = pi/4

x = cos(pi/4)^3 = (sqrt(2)/2)^(3) = 2 * sqrt(2) / 8 = sqrt(2)/4
y = sin(pi/4)^3 = (sqrt(2)/2)^3 = sqrt(2)/4

(sqrt(2)/4 , sqrt(2)/4) is the point